Trigonometry Basics Basic Terms θ (theta) variable for any angle. Hypotenuse longest side of a triangle. Opposite side opposite the angle (θ). Adjacent side next to the angle (θ). Which side is opposite? It depends on the angle. Y is opposite to θ 1 ; Y is adjacent to θ 2. Y θ 2 a R is the hypotenuse for both θ 1 and θ 2. c R b θ 1 X X is adjacent to θ 1 ; X is opposite to θ 2. Pythagorean Theorem Remember that a 2 + b 2 = c 2 Where a and b are either of the sides and c is the hypotenuse. Trigonometric Functions Sin (Sine) ratio of the opposite side to the hypotenuse. Cos (Cosine) ratio of the adjacent side to the hypotenuse. Tan (Tangent) ratio of the opposite side to the adjacent side. Sin, cos, and tan are ratios, telling you how big (what percentage) one side is in relation to another. opp Sin θ = hyp 60º 16 m adj 8 m Cos θ = hyp 30º opp Tan θ = 13.9 m adj Trigonometric Ratios for This Triangle opp. 13.9m Sin60 = = = 0.866 hyp. 16m opp. 8m Sin30 = = = 0.5 hyp. 16m opp. 8m Tan30 = = =.5774 adj. 13.9m Calculators and Trig You will have to find the sin, cos, or tan ratios OR you will have to find θ, given the ratios. You must know how to use your calculator to do this. Given the angle (θ), find sin, cos, or tan. Scientific Calculator: Type 30 Push sin Answer: 0.5 Graphing Calculator: Push sin Type 30 Answer: 0.5 Ex: sin 30 = Answer: sin 30 = 0.5 Given sin, cos, or tan, find θ. Scientific Calculator: Type.4226 Push INV then sin Gives 25 Graphing Calculator: Push 2nd then sin Type.4226 Gives 25 Ex: sinθ =.4226-1 θ = sin (.4226) θ = Answer : θ = 25 Caution! Scientific Calculator: Press the DRG key until degrees shows in the display. Recheck! Use Degrees not Radians! When using sin, cos, and tan your calculator MUST be in degrees or all of your numbers will be wrong. Quick Check: The sin 30º = 0.5! If sin 30º 0.5, your calculator in in radians and must be changed. DRG Graphing Calculator: Press the MODE key. Find where RADIANS is selected. Select DEGREES instead. Press ENTER. Recheck! MODE Finding Unknowns The real power of trigonometry is that it relates the angles and sides of a right triangle. If, for example, you know θ and a side, you can find all the other parts of the triangle. Y Problem: Find the length of X. 60 cm X 25º Step 1: Assign Variables Variables: θ = 25º opp. = Y adj. = X hyp.= 60 cm Step 2: Choose a Formula You know hyp and need adj., so use cos. adj. cosθ = hyp. adj. cosθ = hyp. X cos25 = 60cm Step 3: Solve X.9063 = 60cm (60).9063 = X X = 54.4cm
Trigonometry Basics-2 Using your calculator, give the following ratios. Given the following ratios, use your calculator to find θ. Cos 30º =.8660 Tan 15º = Sin 60º = Tan 30º = Sin 45º = Tan 85º = Cos 45º = Cos 60º = Sin 30º = Tan 45º = Cos θ =.8192; θ = 35º Sin θ = 0.5 θ = Tan θ =.8391 θ = Sin θ =.866 θ = Tan θ = 1.732 θ = Cos θ = 0.5 θ = Tan θ = 1 θ = Cos θ =.866 θ = Sin θ =.7071 θ = What is the sin of 75º? Two sides of this triangle are given. Calculate the third side.? 13 cm If tan θ = 0.868, solve for θ. 12 cm θ = Opposite = Adjacent = Hypotenuse = 9.6 cm 15 cm 11.5 cm 40º θ = Opposite = Adjacent = Hypotenuse = 2 in 75º 8 in 7.7 in Adjacent for 20º = Hypotenuse for 70º = Opposite for 70º = Hypotenuse for 20º = Adjacent for 70º = Opposite for 20º = If opposite = 17.3 cm, then θ = 10 mm 20º 70º 17.3 mm 120 mm Opposite for 30º = Hypotenuse for 60º = Adjacent for 60º = Hypotenuse for 30º = 10 m If opposite = 17.3 cm, then θ = If adjacent = 10 m, then θ = If adjacent = 17.3 cm, then θ = 60º 20 m 30º 17.3 cm Step 1: Assign Variables 26 cm = 35º = Y = Step 2: Choose a Formula (Sin, Cos, or Tan?) Step 3: Solve 26 cm Y 35º Y = Y sin30 = 10cm Y.988 = 10cm Y = 9.88cm Obviously Y 9.88 cm. What went wrong? Following the three steps at the left, find the length of X. 10 cm Y 30º 6 ft 45º X How long is Y?
Vector Basics We use arrows to represent vectors because vectors have both magnitude (size) and direction (which way it points). 7 m/s 45 Same magnitude; different directions. 7 m/s -30º 6 m -30º 4 m Same direction; different magnitudes. 2 kg Mass is a scalar, which only requires magnitude. You don t need the direction of the mass. Adding Graphically When adding vectors graphically, put the arrows head to tail. The resultant points from start to finish. In this example your total displacement is the straight line distance between your initial and final position NOT the distance you traveled. The result of adding together two or more vectors is called a resultant. Resultant Total displacement start final Vector 1 Displacement 1 Vector 2 Displacement 2 Order doesn t matter when adding vectors. The resultant will be the same. V 1 V 1 V 2 finish start V 2 R 1+2 R 1+2 finish start V 1 Different order: same resultant. V 2 40 m 30 o x component = 34.6 m x o cos 30 = 40 m x = (40m)cos 30 x = (40m )(.866) = 34.6 m +y component Components o The components are the portions of the vector in the x or y direction, like coordinates on a graph. y component = 20 m y o sin 30 = 40 m y = (40m)sin 30 y = (40m)0.5 = 20 m The components tell you that you would have to move 34.6 m in the x-direction and 20 m in the y-direction to move 40 m at 30º Components can be negative or zero. 100% of this vector is vertical 90 o No x comp. +y component Use 130 and your calculator will give you the correct + and components. 75 m 50 o -x component 130 o An example of no x-component: a cat climbs up a tree. The cat doesn t move horizontally, just vertically. o Think directions NOT angles. Your calculator will give you positives and negatives automatically IF you give the calculator correct directions. These angles are equal (20º), but point in different directions. Setting up vectors. Before you calculate components always find direction of θ, as shown above. X and Y Components: 180º 180 + 20 = 200º 15m 30º 180º X-comp = Hyp.(cosθ) 90º -90º 90 + 20 = 110º 20º 20º 90º 270º or -90º 0º For your calculator only this is 20º. 20º 20º Direction of the vector 0º For your calculator this is -20º. Y-comp = Hyp.(sinθ) Magnitude (length) of the vector The angle is 30º, but the direction is 150º (180 30). This vector is 15 m at 150º Units Components retain the units of their vector (and vice-versa). 30 o vector = 6 m/s x component = 6m/s(cos 30 o ) = 5.2 m/s y component = 6m/s(sin 30 o ) = 3 m/s If the vector was a plane, think of the x-component as a race car trying to stay beneath the plane on the ground. The y-component could be how fast the plane gains altitude. Math and Vectors Subtracting vectors: add its opposite (the negative of the vector). Opposite of V 1 V 1 V 1 Multiplying vectors: multiply the size of the vector. Twice the size of V 1 V 1 + V 1 = 2V 1
Vector Basics -2 1. Resolve 2. Magnitude 3. Resultant 4. Component A. The portion of the vector on the x or y axis. B. To find the x or y-component of a vector. C. The size of a vector ( 35 of 35 m ). D. Tells where a vector is pointing or the angle of the vector. 7. In figures A D, which vectors are added correctly? If wrong, why? A. B. R A. V 2 V 1 R B. V 2 V 1 5. Direction 6. Vector E. What you find by adding two vectors together. F. Something that has magnitude and direction. C. D. R C. V 2 V 1 R D. V 2 V 1 Using the vectors at the right, draw the resultants for the following operations. V 1 V 2 V 3 V 4 V 5 8. V 2 V 5 = 10. V 3 + 2V 4 V 5 = 11. 2V 1 2V 4 = 9. 2V 2 + V 4 = 12. Add 2V 1 +V 4 mathematically. 13. If each of the vectors is 10º from the closest axis, determine the directions of each of the vectors. θ A = θ B = θ C = θ E = θ F = θ G = D 180º E 90º C B A 0º H 15. A person walks 12 m across a room. A. What is their horizontal component? B. What is their vertical component? 16. Resolve this vector into its components. θ D = θ H = F G 270º or -90º 20 m/s 14. Find calculator directions for the following vectors. A. θ = B. θ = C. θ = 30 o 25º 45º 17. Resolve these vectors into their components. A. 42 m/s 35 o B. 20 m/s 2 45 o C. 70 o 72 m
Period: Each grid square represents 1 m. In this example, you may count squares. Ex 1 finish 1. Find the following information for vector 1. A. How far does vector 1 move horizontally? (This is the X-component.) X 1 = V2 = B. What is the Y-component of vector 1? Y 2 = C. How long is vector 1? (Find the magnitude of vector 1.) start V 1 = 2. Resolve vector 2 into its x and y components. (Do the same as in #1) A. X 2 = B. Y 2 = C. Magnitude of vector 2 = 3. Draw the resultant from the start to the finish (Label it R ). 4. Add together X 1 and X 2 = (this is X total ) 6. Using X total and Y total, calculate the length of R (with X total and Y total you have two sides of a right triangle). 5. Y total = Each grid square represents 1 m. In this example, you may count squares. Ex 1 finish 1. Find the following information for vector 1. A. How far does vector 1 move horizontally? (This is the X-component.) X 1 = V2 = B. What is the Y-component of vector 1? Y 2 = C. How long is vector 1? (Find the magnitude of vector 1.) start V 1 = 2. Resolve vector 2 into its x and y components. (Do the same as in #1) A. X 2 = B. Y 2 = C. Magnitude of vector 2 = 3. Draw the resultant from the start to the finish (Label it R ). 4. Add together X 1 and X 2 = (this is X total ) 5. Y total = 6. Using X total and Y total, calculate the length of R (with X total and Y total you have two sides of a right triangle).
Ex 2 1. A. X 1 = B. Y 2 = C. Magnitude of V 1 = finish 2. A. X 2 = B. Y 2 = C. Magnitude of vector 2 = V 2 = start V 1 = 3. Draw the resultant from the start to the finish. 4. A. X total = B. Y total = 5. Calculate the magnitude of R. 6. Using X total and Y total, calculate the direction of R. (see Adding Vector notes.) Ex 2 1. A. X 1 = B. Y 1 = C. Magnitude of V 1 = finish 2. A. X 2 = B. Y 2 = C. Magnitude of vector 2 = V 2 = start V 1 = 3. Draw the resultant from the start to the finish. 4. A. X total = B. Y total = 5. Calculate the magnitude of R. 6. Using X total and Y total, calculate the direction of R. (see Adding Vector notes.)
Adding Vectors Setting Up Individual Vectors Before you begin, be sure all of your vectors have the same units and all angles start at the +x-axis. Then, your calculator will automatically calculate any positive or negative components. y o sin120 = 45 m o 45(sin120 ) = y 45(.866) = y y = 39 m y = 39 m x o cos120 = 45 m o 45(cos120 ) = x D = 45 m 60 o x = -22.5 m X and Y Components x = Hcosθ and y = Hsinθ Where H is the vector, θ is the angle from the + x axis, x is the x component of H, and y is the y-component of H. From + x axis 120 o +x 45(-0.5) = x x = -22.5 m All angles MUST start at the + x axis Using 120 gave a x component! These 2 equations work for any vector (even if the vector is vertical [θ = 90º or 270º] or horizontal [θ = 0º or 180º]). Example: A person walks 35 m at 30 then 50 m at 60. Calculate the person s total displacement. Step 1. Resolve vectors into their components. R can be found graphically (by drawing and measuring) or mathematically. 60 o Vx 2 = 25 m 30 o 17.5 m Vy 1 = Vx 2 = 25 m o Vx 1 = 30.3 m Vy 1 = 35(sin30 ) o Vx = 35(cos30 ) Vy 1 = 17.5 m 1 1 v 1 = 35 m Vx = 30.3 m v 2 = 50 m Resultant (total displacement) Vy 2 = 43.3 o Vy 2 = 50(sin60 ) Vy = 43.3 m 2 o Vx 2 = 50(cos60 ) In this example all components are positive. By taking your angle from the +x-axis, sine and cosine will give you positive and negative components automatically. Step 2. Calculate x total and y total by adding up all x-components and all y-components. Step 3. Draw a resultant triangle with x total and y total. Then, calculate the resultant s magnitude and direction. Be sure to keep track of negatives! Resultant (total displacement) x 1 y 1 x total = x 1 + x 2 x 2 = 30.3 + 25 = 55.3 m y 2 ytotal = y1 + y2... = 43.3 + 17.5 = 60.8 m Magnitude (R) = = 82.2 m θ = 47.7 o 2 2 total total x + y Direction = -1 y total θ = tan x total x total = 55.3 m Answer: R = 82.2 m at 47.7º ytotal = 60.8 m Use the Pythagorean Theorem to find the resultant s magnitude. R = x + y 2 2 2 total total 2 2 2 R =55.3 + 60.8 = 6754.73 R = 6754.73 = 82.2 m Use inverse tangent to find the resultant s direction. opp. y tan θ = = adj. x 60.8 tan θ = 55.3 60.8 55.3 total total -1 θ = tan = 47.7 Tan Can t See X -4 = ( ) = 2 Tan sees these as the same! 1 1 tan tan 2 63.4 4 = ( ) = -2 1 1 tan tan 2 63.4 Tan can t see the difference between a negative x and a negative y. Tan only gives angles between +90 and 90., so it will never give you an angle in quadrants 2 or 3. Ytotal = 10.4 m θ X total = -6 m 90º If X total is negative add 180 degrees 1 10.4 tan = 60-6 0º Since X total is negative: θ = 60 + 180 = 120º But we know that this direction is greater than 90º. Which we can see is true.
Adding Vectors -2 1. A drag racer moves 350 m down a race track. X = Y = 3. Resolve the following vectors into their components. 2. A person walks 150 m north. X = Y = A. 6.5 m 32º B. 40º 12 m/s C. 25º 80 m 4. Find the resultant of the following two vectors. 5. Add these vectors together. Assume each square = 1 m. start finish Y 2 = 3 m X 2 = 5.2 m V 1 = 12 m 40º V 2 = 6 m 30º X 1 = 9.2 m Y 1 = 7.7 m D. Calculate the magnitude (length) of R. A. Draw the resultant from the start of the first vector to the end of the second. Label it R. B. X total = C. Y total = E. Calculate the direction of R. V 1 V 2 A. X 1 = Y 1 = B. X 2 = Y 2 = C. X total = Y total = D. Draw the resultant (R). E. Calculate R s magnitude. F. Calculate R s direction. 6. A person walks 30 m north, then 50 m at 35º. Find their total displacement. A. Below draw R from the start of V 1 to the end of V 2. B. Resolve v 1 and v 2 into their components (Step 1 on the front) V1 = 30 m V 2 = 50 m 35º X 1 = X 2 = X total = Y 1 = Y 2 = Y total = Magnitude = Direction = 7. Add these vectors: V 1 = 55 m at 38 o and V 2 = 10 m at 260 o. A. Draw V 1, V 2, and R below. B. Resolve v 1 and v 2 into their components (Step 1 on the front) X 1 = X 2 = X total = Y total = Y 1 = Y 2 = Magnitude = Direction =
Additional Adding Vector Examples Ex. 1 Add these vectors: 35 m at 30 and 60 m at 160. These additional examples are for students to visualize the process of adding vectors. It is assumed that students can already calculate components. Step 1. Resolve vectors into their components. Step 2. Calculate x total and y total y 2 = 20.5 m x 2 = -56.4 m D 2 = 60 m D 1 = 35 m 30 o x 1 = 30.3 m 160 o y 1 = 17.5 m ytotal = y1 + y2... = 20.5 + 17.5 = 38 m y 2 R x total = x 1 + x 2 x 2 = 30.3 56.4 = 26.1 m x 1 y 1 Step 3. Calculate R s magnitude and direction. R = x + y 2 2 2 total total R =26.1 + 38 2 2 2 R =46.1 m ytotal = 38m x + y R = 46.1 m 2 2 total total θ = tan θ = 124.4 y x -1 total total opp. y tan θ = = adj. x total total tan θ = 38 26.1 38-26.1 56.6 + 180 = 124.4-1 θ = tan = 56.6 Add 180 to your angle because it is obviously not in the 4th quadrant. x total = 26.1m Ex. 2 Add these vectors: 1250 m/s at 90 and 2600 m at 320. Step 1. Resolve vectors into their components. Step 2. Calculate x total and y total θ 2 = 320º x 2 = 1992 m/s x 2 1250 m/s at 90 y 1 = 1250 m/s 2600 m/s at 320 y 2 = 1670 m/s y 1 x total = x 1 + x 2 = 1992 y 2 R R y total = y 1 + y 2... = 1250 1670 = 421 m Step 3. Calculate R s magnitude and direction. R = x + y 2 2 2 total total R = 1992 + 421 2 2 2 R =2036 m x t = 1992m R y t = 421 m opp. y tan θ = = adj. x 421 tan θ = 1992 421 1992 total total -1 θ = tan = 12
Relative motion describes an object s motion relative (in relation) to different frames of reference. Relative Motion V truck to road = 5 m/s V tortoise to truck = 0.1 m/s In the picture at the right the tortoise has a speed of 0.1 m/s relative to the truck, but has a speed of 5.1 m/s relative to the road. In describing relative motion you must always give your frame of reference (the truck or the road). V tortoise to road = 5.1 m/s Vectors Still Add Relative motion can be solved graphically or mathematically, just like all other vectors. V Wind = 20 m/s 45º V Plane = 40 m/s A plane flying has a velocity of 40 m/s, relative to the ground, experiences a west wind that is moving 20 m/s, relative to the ground. What is the velocity of the plane relative to V Wind = 20 m/s the ground? Graphically: V Plane = 40 m/s V total Vx wind = 20 Vx plane = 40(cos45º) = 28.3 Vx total = 20 + 28.3 = 48.3m/s Vy wind = 0 (all horizontal) Vy plane = 40(sin45º) = 28.3 Vy total = 0 + 28.3 = 28.3m/s Mathematical Solution: See notes: Adding Vectors V total V = total 2 2 48.3 + 28.3 = 56 m/s 28.3 θ = tan-1 48.3 θ = 30.4 X and Y are Independent 2 m V train = 0 m/s V person = 1 m/s end start V person start To calculate V total, use Pythagorean theorem: V total 2 = V person 2 + V train 2 V train V total end Imagine a person walking 1m/s across a 2 m wide railroad car. If the car is at rest the person will take 2 seconds to cross. If the railcar is moving, the person still takes 2 seconds to cross the car. The velocity of the railcar is irrelevant because the velocity of the car is in the x-direction and the person is moving in the y-direction. The x and y directions are independent of each other! Not all motion is possible Remember that the resultant is drawn from the start of the first vector to the end of the last vector. In the example below, a boat encounters a strong current in a river. If the boat captain aims straight across the river, the river will push the boat downstream. But what if the captain wishes to land at a point straight across the river? The boat will have to be aimed up stream, but at what angle? River 7 m/s launch Boat 5 m/s V B V R V T landing landing River 7 m/s Boat 5 m/s launch Desired path V B = 5 m/s V R = 7 m/s V T The total velocity of the boat is found by adding the vectors, which here requires only Pythagorean theorem, because these vectors are already horizontal and vertical. A vector triangle shows that this is not possible because the hypotenuse is smaller than one of the sides. The boat is not able to go straight across.
1. An moving walkway at the airport has a velocity of 2 m/s to the right. A person walks at a steady pace of 3 m/s. A. If the person is walking to the right, what is their velocity relative to the walkway? B. What is their velocity relative to the ground? C. How long would it take them to travel to the food court, 100 m away? 2 m/s Relative Motion -2 3 m/s E. How long would it take them to walk to the food court and back without using the walkway? D. How long would it take them to walk back if they have to walk on the same walkway? 2. A toy plane s is flying 55 going 8 m/s. If the wind is pushing with a velocity of 3 m/s at 30º, find the total velocity and direction of plane. 3. A boat is traveling 6 m/s at an angle of 30 o. The water has a current flowing 3 m/s directly south. Find the boat s total velocity and direction. 4. A person can swim 4 m/s. The river has a current flowing 6 m/s directly east. A. What will be the direction and velocity of the person if they aim directly across the river (north)? B. If the person swims at constant speed, how long does it take them to swim across the 40 m wide river? C. How far downstream will the person drift? D. At what direction will the person have to swim to reach a point directly across the river? E. If the river s current increases (is faster), will the person take more or less time to cross the river?
Hypo. can t be smaller than a side. Even if they swam directly upstream they would be pushed downstream. They have to be swimming faster than the river to get straight across. Then the hypo is bigger than both sides. The x and y directions are independent of each other. The x-direction speed of the river has no effect on a person swimming in the y-direction.
Projectile Motion A projectile is any thrown, shot, or launched object: a rock; a bullet; a volleyball; a person jumping. All projectiles follow a parabolic path. The distance a projectile travels in the x-direction is known as its range. If you took several quick pictures of a projectile in flight you would notice that in the y-direction it looks just like freefall being pulled back to the earth by gravity. In the x-direction you would see that the object is at constant speed, moving the same distance each second. In the y-direction it is in freefall. So a y = 9.8 m/s 2. Vy i = V i sinθ θ V i Vy Vx i = V i cosθ Vx i = V i cosθ Vx At the top: V y = 0 m/s, so V = Vx The object stops in x-direction when it hits the ground in the y-direction, so: t y = t x Ground To find the projectile s velocity and angle at any point: calculate Vx and Vy, then use Pythagorean Theorem and inverse tan. stops In the x-direction the object is at constant speed. So a x = 0 m/s 2 and S = D/T. Vy f = Vy i Just as with any vector, the x and y-components of the initial velocity can be found using sine and cosine. (Initial vertical velocity) Vy i = V(sinθ) Initial velocity (velocity it was thrown or shot) V i θ Vx i = V(cos θ) (Initial horizontal velocity) Vy i = V(sinθ) 6 m 3 m Time Comes From the Y-direction An object launched from higher up stays in the air longer and goes farther. Vy i = 0 m/s V i = Vx i If an object is launched horizontally, the initial vertical velocity is 0 m/s. You know that if you throw a rock from up on a cliff, it will go farther than if you were on the ground. This is because the higher up you are, the longer the rock is in the air. Flight time is dependent on the y-direction only! Solving Projectile Motion Problems Any projectile motion problem can be easily solved if you : 1) draw the situation; 2) write out what you know in the x and y-directions; 3) solve for unknowns. Use your Freefall notes to help you assign the y-direction variables. Just as in all freefall problems, you will use one of the kinematic equations to calculate unknowns. Often you will solve for time in the y-direction. 2 y direction a y = 9.8 m/s 2 V i = 0 m/s V f = not used y = 8 m t =? 1 2 y = ( vit) + a( t) 2 1 2 8 = (0 t) + ( 9.8)( t) 2 t 8 = 4.9t 8 = = 1.63 4.9 t = 1.63 = 1.28 sec 2 9 m/s 8 m The accelerations will ALWAYS be 9.8 m/s 2 and 0 m/s 2. Since horizontal: Vy = 0 m/s x (range) =? x direction a x = 0 m/s 2 V i = 9 m/s V f = 9 m/s x = t = 1.28 sec D S = T D = ST D = 9(1.28) D = 11.5 m Because it s at constant velocity From the y-direction Since the x- direction is at constant speed, you can ALWAYS use S = D/T for your x-direction equation.
Problem: A cannon fires a cannonball. If the cannon s angle is 35 o to the ground and the muzzle velocity is 50 m/s, what is the range of the cannonball? Step 1: Resolve V into Vx and Vy Vi VYi VYi = θ V(sin θ)= θ = 35 o 28.7m/s VXi VXi = V(cos θ) = 41 m/s Projectile Motion Example Vi = 50 m/s θ = 35 o landing X =? Vi = 50 m/s X-component VX = V(cos θ) VX = 50m/s(cos 35 o ) VX = 41m/s Y-component VY = V(sin θ) VY = 50m/s(sin 35 o ) VY = 28.7m/s Step 2: Find time from the y-direction. In the y-direction, the cannonball is in freefall. Write the variables you know: ay = 9.8 m/s 2 Vi = Vy = Vsinθ = 28.7m/s (see step 1) Vf = -Vi = -28.7m/s Y = 0m (from ground to ground) t =? Choose one of the kinematic equations: 1 y = ( v i + v f ) t 2 v = v + a t f i ( ) 1 2 2 y = v t + a t 2 y = v t at 2 2 f i 1 2 ( 2 ) v = v + a y Step 3: Find x from the x-direction. In the x-direction, the cannonball travels at constant velocity: Vx, the x-component of the velocity. The time of flight we just found in the y-direction. Write the variables you know: ax = 0 m/s 2 Vi = Vf Vx = Vcosθ = 41 m/s (see step 1) t = 5.8 sec (from y [see step 2]) x =? S = D T S = D T D = S T D = 4 1 ( 5.8 ) x = 237.8 m By knowing only the launch velocity and angle, you can calculated that the cannonball traveled 237.8 meters in 5.8 seconds. i f Projectile Motion -2 Since you have all of the variables, you can use any formula. So use the easiest one: vf = vi + (at) Put in what you know: -28.7 = (28.7) + (-9.8t) 28.7 28.7-57.4 = -9.8t by 9.8 t = 5.8 sec Since ax = 0 m/s 2 RANGE This is the most basic of all projectile motion problems: from the ground-to-theground (A to E). Steps 1 and 3 will ALWAYS be the same. Step 2 is just freefall and will change according to the situation. Muzzle velocity
Projectile Motion Concepts Total Speed A projectile s speed at any point is the vector sum of Vx and Vy at that point (use Pythagorean Theorem). Only if you are asked for the vertical or horizontal speed do you use one of the components. Vertical Speed Vy = 28.7 m/s V = 50 m/s 35 Vx = 41.0 m/s Horizontal Speed 2 2 Total Speed = Vx +Vy Total Speed Other words: Speed Total velocity Initial Velocity Velocity Vy = V= 0 m/s Vy i = V i = 5 m/s Since this projectile is thrown straight up (at 90 ), its speed (V) is equal to the vertical speed (Vy) at all times. Y-direction Questions How high? How much time How long will it be in the air? - Since the y-direction is freefall, being pulled down by gravity, these questions are only answered in the y-direction. A 12 m/s B 6 m/s v y = 17.1 m/s v = 50 m/s v y = 30.3 m/s v = 35 m/s C 20 v x = 47.0 m/s 60 D v x = 17.5 m/s Since they are at the same height, both balls will hit the ground at the same time, but Ball A will go twice as far in that time. Object C has the greater initial speed, but object D will go higher and take longer to get back to the ground because D has the greater vertical speed. y position (m) Greatest Range 100 90 80 70 60 50 40 30 20 10 0 Same Same Initial Velocity, Speed, Different Different Angles Angles 70 20 How you determine which projectile has the greatest range depends on whether the projectiles have the same initial velocity. 35 65 45 0 20 40 60 80 100 120 140 160 180 200 x position (m) If projectiles are shot with the same velocity, but different angles, the range increases the closer the launch angle gets to 45. Notice that 20 and 70 hit at the same point and both are 25 away from 45. Also, notice that 70 goes the highest because it has v y = 17.1 m/s f C Different Initial Speeds 20 v = v + at 17.1 = 17.1 9.8t 34.2 = 9.8t t = 3.5 sec i v = 50 m/s v y = 47.0 m/s D S = T D = ST D = 47.0(3.5) D = 164.5m v y = 30.3 m/s v = v + at v = 35 m/s 60 D v x = 17.5 m/s 30.3 = 30.3 9.8t 60.6 = 9.8t t = 6.18 sec Even though 60 is closer to 45 than 20, we cannot assume that object D will have the greatest range, since the balls do not have the same initial speed. Instead, we must calculate the range as with any ground-to-ground projectile problem (calculate time in y-direction and range in the x-direction). This reveals that object C actually has the greater range. f i D S = T D = ST D = 17.5(6.18) D = 108.2m
Projectile Motion Special Situations All projectile motion problems work the same. First you resolve the initial velocity into Vx i and Vy i. Second, you write everything you know in the x and y-directions. Third, remembering that t y = t x (times are the same in both directions), you solve. This, of course, assumes that you know the basics, such as a y = 9.8 m/s 2 and a x = 0 m/s 2, etc. Horizontal Launch For any horizontally launched object, θ = 0º, Vx i = V and Vy i = 0 m/s. 2 m/s 4 m Example 1: A ball is shot 2 m/s horizontally from 4 m up. How far away will it land? You should already know: a y = 9.8m/s 2 a x = 0m/s 2 y = 4 m Vx = 2 m/s Vy i = 0 m/s As always, find time in the y-direction: Variables: 1 2 y = ( vit) + ( at ) a y = 9.8m/s 2 2 y = 4 m 1 2 4 = (0 t) + ( ( 9.8) t ) Vy 2 i = 0 m/s 2 t = 4 = 4.9t 2 Vy f = not used.82 = t t = 0.9 sec D S = T D ( or x) = ST x = 2(0.9) x = 1.8 m No acceleration in the x-direction, so use the easy equation! Maximum Height Find the maximum height or how high? : these are purely y-direction questions, so the x-direction can be ignored. Example 2: A ball is shot 58 m/s at 40º. How high up will it go? How high? 58 m/s 40º Step 1: Find Vy i. 58 m/s 40º Vy i = 58sin40º = 37.3 m/s Write what we know in the y-direction and solve: Variables: a y = 9.8m/s 2 y = Vy i = 37.3 m/s Vy f = 0 m/s (at the top) t = not used V = V + (2 a y) 2 2 f i 2 0 = (37.3 ) + (2( 9.8) y) 0 = 1391.29 + ( 19.6 y) 1391.29 = 19.9 y 1391.29 = y 19.9 71 m = y As with a ground-to-ground example, these two special situations work the same way every time. More importantly, though, is for you to see the commonality of all projectile motion problems so that you can solve new problems, when they arise.
Projectile Motion Concepts Assign variables for each of the following situations. Write? for any unknown. 1. An object is shot 8 m/s horizontally from a 7 m tall cliff. How far away does it land? y-dir. a y = Vy i = Vy f = y = t = 3 sec x-dir. a x = Vx i = Vx f = x = t = 2. An object is shot 40 m/s at an angle of 50º. How high does it go? y-dir. a y = Vy i = Vy f = y = t = 6.9 sec x-dir. a x = Vx i = Vx f = x = t = 3. An object is shot 80 m/s at 60º. Calculate range. y-dir. a y = Vy i = Vy f = y = t = x-dir. a x = Vx i = Vx f = x = t = 2.5 sec
4. An object is shot 80 m/s at 60º. Find its maximum height. y-dir. a y = Vy i = Vy f = y = t = 2 sec x-dir. a x = Vx i = Vx f = x = t = 5. An object is shot 40 m/s at an angle of 50º. How far away does it land? y-dir. a y = Vy i = Vy f = y = t = x-dir. a x = Vx i = Vx f = x = t = 6 sec 6. An object is shot 8 m/s horizontally from a 7 m tall cliff. Find its range. y-dir. a y = Vy i = Vy f = y = t = 3 sec x-dir. a x = Vx i = Vx f = x = t =
Projectile Motion Concepts with Diagrams 1. An object is thrown into the air going 80 m/s at an angle of 60º. How high does it go? A. Realizing that in the y-direction projectiles are just freefall, fill in the y-direction variables. B. Realizing that in the x-direction, projectiles are at constant speed, fill in the x-direction variables. C. In the y-direction, calculate how high the object goes. 2. An object is launched horizontally with a speed of 8 m/s. A. Since it is launched horizontally, what is the initial y-direction velocity? B. What is its initial x-direction velocity? C. Again, in the y-direction projectiles are just freefall, fill in the y-direction variables. D. In the x-direction, projectiles are at constant speed, fill in the x-direction variables. F. In the y-direction, calculate how much time it is in the air before it hits the ground. G. In the x-direction (at constant speed), what equation will you use? H. Calculate how far away it landed in the x-direction, using the time you just found. 3. An object is shot 40 m/s at an angle of 50º from the ground. How far away does it land? A. Fill in the x and y variables for the object. B. Calculate how long it was in the air, in the y-direction. C. In the x-direction (at constant speed), use the time you just calculated to find how far away it landed.
4. An object is shot 50 m/s at an angle of 70º. How high does it go? A. Use trigonometry to calculate the initial x and y velocities of the object. B. Fill in the x and y variables. C. Calculate how high the object rises. 5. An object is launched 15 m/s horizontally. A. Fill in the variables for the object. B. Solve for time in the y-direction. C. Since the x-direction is constant speed, solve for x. 6. An object is launched from the ground at a speed of 30 m/s at an angle of 55º. If it lands back on the ground, calculate how far it went horizontally. A. Find the initial x and y velocities from the given speed and direction. B. Fill in the variables. C. Calculate time in the y-direction. D. Calculate x.
Projectile Motion Comparisons 1.Which one hits the ground first? 2. Which has the greater maximum height? 3. Which one has the greatest range (lands farthest from the table)? 4. Which one has the least range (land closest to the table)? 5. Which as the greater maximum height? 6. Which one has the greatest acceleration when it leaves the table? 7. Which has the greater maximum height? 8. Which one hits the ground last? 9. Which has the greater maximum height? 10. Which has the greater speed at the top of its path? 11. Which has the lesser maximum height? 12. Which one hits the ground first? 13. Which one has the greater vertical displacement? 14. Which one has the smaller maximum height? 15. Which has the greater vertical acceleration? 16. Which one has the greater range? 17. Which has the greater acceleration? 18. Which has the greater range?
Projectile Motion Comparisons -2 19. Which has the greater horizontal velocity? 20. Which one hits the ground first? 21. Which has the greater horizontal acceleration? 22. Which one is in the air for less time? 23. Which one has the greatest vertical acceleration at the top of its path? 24. Which has the smaller range? 25. Which is in the air for the most amount of time? 26. Which one has the greater maximum height? 27. Which has the greater range? 28. Which one has the greatest range? 29. Which as the greater maximum height? 30. Which one has the greater velocity at the top of it s path? 31. Which one has the greatest maximum height? 32. Which one is in the air for the least amount of time? 33. Which one has the greater range?
Satellites and Review Imagine a ball thrown at the top of a hill. Because the ground falls as the ball falls, the ball goes farther. If you were able to throw the ball fast enough, the earth would curve away from the ball as fast as the ball falls, causing it to be in orbit around the earth. It would be a special projectile: a satellite! Satellites are still falling toward the earth, but they are going fast enough to stay in orbit. For dropped or thrown objects Vy i = 0 m/s and a y = -9.8 m/s 2. So, in 1 second, it will drop: 1 2 y = (0 t) + ( 9.8)(1) 2 y = 4.9(1) = 4.9m 1 sec = 4.9 m 8,000 m (not to scale) The earth s curvature takes about 8,000 m to falls 4.9 m. To be a satellite, an object must go 8,000 m in the 1 second it falls or 8,000 m/sec (8 km/s) or 18,000 mph. 4.9 m Review 1. Resolve 2. Magnitude 3. Resultant 4. Component 5. Direction 6. Vector A. 30 of 35 m at 30. B. 35m of 35 m at 30. C. How much of the vector is in the x or y direction. D. Using trigonometry to finding the x or y-components of a vector. E. The total from start to finish after adding vectors together. F. Any physical quantity that requires both magnitude and direction. 7. Scalar or vector? A. A person drives 35 m/s. B. Velocity C. Mass of a person. D. acceleration 8. A cat climbs 3 m up a tree. Resolve this into its x and y components. 9. A = 10. B + = 0 11. A + D = 12. D = 2D A C E G 13. Which vector/s have +X and Y components? 14. Which vectors have no x components? B D F H 15. Which vectors have equal x and y components? 16. Mathematically, what is A + H D + 2E? 17. Draw 2E G +F C. 18. On the diagram below give two ways to make R. C A R D B
Satellites and Review -2 19. What is a x? 20. If Vx i = 12 m/s, what is Vx f? 21. So, how do Vx i and Vx f compare? 22. What is a y? 23. If a projectile is launched from the top of an 9 m tall building, what is y? 24. How do you calculate Vy i? 25. How do you calculate Vx i? 26. In which direction do you calculate time? 27. In which direction is a projectile freefall? 28. In which direction is a projectile in constant motion? 29. On the diagram at the right draw Vx and Vy at each point on the projectile s path. Vy B C D A Vx E 30. Which one will be in the air the longest? 31. Which one will have the greatest range? A 5 m/s A 5 m/s 5 m/s 5 m/s B 4 m/s B 5 m/s 32. Which one goes higher into the air? 33. Which one has the greatest acceleration? 34. Which one has the greatest range? 35. For a projectile shot from the ground to the ground which angle gives the longest range? 36. Which angle gives a greater range: 20 or 30? 37. Which angle gives a greater range: 40 or 50? 38. Which angle gives a greater range: 20 or 70? 39. A projectile is shot 25 m/s at an angle of 34. How high up into the air does the projectile go? A. Find Vy. C. Solve for how high. B. Write the y-direction variables. 40. A person can swim 2 m/s in still water. The river they are swimming in has a current of 1 m/s. A. How fast are they swimming relative to the water if they swim with the current? B. How fast are they swimming relative to the shore if they swim with the current? C. How fast are they swimming relative to the shore if they swim against the current? D. If they swim straight across the river how fast are they swimming relative to the shore? V person = 1 m/s V total =? V river = 1 m/s
Two dimensional motion is not as difficult as it may seem if you understand the logic. This flow chart helps you ask the right questions and refers you to the appropriate reference material (worksheets) to understand any two dimensional problem. START Two Dimensional Motion Are the units the same for all quantities (all m/s, for example)? No Yes Resolve velocity into components No Is the acceleration (force) in the same direction as the velocity? See worksheet: Projectile Motion x direction Place Vx and Vy into x and y directions y direction Yes Use the kinematic equations at that angle (this is the hypotenuse) Fill in x-direction variables, remembering that ax = 0 m/s 2 Use special situations on Free-fall worksheet to fill in y-direction variables Solve for unknown quantities Solve for any unknown quantities BIG HINT: Time is the connection between the x and y directions Solve for any unknown quantities Resolve into components if necessary (to find altitude [y], how fast east [Vx], etc) Note: This chart assumes that all situations in which V and a are in different directions are projectile motion. Other possibilities exist (a force pushing an object off course, for example), but they are less common. See worksheet: Adding Vectors Is the vector vertical or horizontal? No Resolve vector into x and y components Label x or y (remember + or -) Add the all x s and all y s into xtotal and ytotal Find magnitude with Pythagorean theorem Find direction θ = tan -1 (yt/xt) Yes