Projectile Motion. v a = -9.8 m/s 2. Good practice problems in book: 3.23, 3.25, 3.27, 3.29, 3.31, 3.33, 3.43, 3.47, 3.51, 3.53, 3.

v a = -9.8 m/s 2 A projectile is anything experiencing free-fall, particularly in two dimensions. 3.23, 3.25, 3.27, 3.29, 3.31, 3.33, 3.43, 3.47, 3.51, 3.53, 3.55 Projectile Motion Good practice problems in book: 3.23, 3.25, 3.27, 3.29, 3.31, 3.33, 3.43, 3.47, 3.51, 3.53, 3.55

Practice test Reminder: test Feb 8, 7-10pm! Email me if you have conflicts! Treat this practice test like the exam. 3 hours, closed book, non-graphing calculator, pencil and paper. This is your next homework (not graded). Do it ASAP - it will indicate how well you ll probably do on the exam. I m not going to ask you to turn it in. While you guys are passing these around I m going to give a few announcements.

Your intuitive understanding of the Physical world Percent grasp of general Physics principles Note: all of you got an A. Tested your perception of basic Physics principles. We re going to work on fixing your misconceptions! For instance: everything falls at the same rate! Forget for a moment that I m grading you

Please ignore your ecampus grade for now! Please ignore the grades posted there until I get things fixed. They might not make sense.

Today! Figure 1. Dysfunctional projectile Decomposing X and Y movements. How to hit a target. How to go about Projectile Motion problems.

Ignoring air resistance, what would be the path of motion if someone ran off of a cliff? Q22 A B C Straight out for a little while then gradually down. D Straight down almost immediately Straight down after some time Never completely straight down Answer: C. Gravity always acting once in free-fall, and has forward motion with nothing pushing it back.

Three velocity vectors. v vx vy Let s think about this guy running off the cliff. And let s consider that he has THREE VELOCITY VECTORS that we need to consider. There s the net velocity he s going in a given direction, there s the horizontal component vx (how fast is he moving parallel to the ground), and there s the vertical component, vy (how fast is he going toward the ground).

Three velocity vectors. v vx vy v = vx0 Starts with horizontal velocity vx0 And vertical velocity vy0 = 0 m/s. When he starts, it s all in the horizontal component. There s no vertical component.

t = 0s Free body! v = vx0 ay = -9.8 m/s 2 This is just as true the very moment he leaves the cliff face. Now the moment he leaves the cliff, he becomes a free body; the only thing acting on him is the force of gravity. This means he gets an acceleration vector downward at good old 9.8m/s^2. I want you to take 30 seconds and tell me something very fundamental here.

t = 0s Free body! Q23 v = vx0 ay = -9.8 m/s 2 Throughout this fall, what is the jumper s horizontal acceleration vector, ax? A. vx0/t B. 9.8 m/s 2 C. 0 m/s 2 D. Not given, so unknown Answer here is C.

t = 0s Free body! v = vx0 ay = -9.8 m/s 2 Acceleration needed to change velocity vx = v x0 + axt In free bodies, the only acceleration is in the y direction vx is always equal to vx0 (unless it hits something) You need an acceleration to change your velocity. This guy, LIKE ALL PROJECTILES, is a free body. Thus, he gets affected by the force of gravity in the Y DIRECTION, this means that over the course of his journey, this jumper s Y VELOCITY VECTOR will gradually change. This is why he falls, and falls faster and faster. However, there is nothing pushing him forward or backward in the x direction, so his horizontal velocity will never change!

v vx vy t = 0s Free body! vx = vx0 t = 1s t = 2s vx = vx0 vx = vx0 t = 3s vx = vx0 Let s look at this in terms of vector decomposition of his movement. I m showing his true velocity as the black arrowno matter how much time goes by, velocity s x component never changes because there s nothing to change it! The guy keeps moving forward. But what about the y component?

v vx vy vy0 = 0 m/s +y t = 0s t = 1s How far downward has the jumper fallen after 3 seconds? t = 2s v = v 0 + at Δx = v 0 t + ½ at 2 v 2 = v 0 2 + 2aΔx t = 3s Answer: B But what about y? Y is changing only due to gravity so we re solving this the SAME WAY WE HAVE BEEN! Using the plain old motion equations.

v vx vy vy0 = 0 m/s t = 0s t = 1s +y t = 2s v = v 0 + at Δx = v 0 t + ½ at 2 v 2 = v 0 2 + 2aΔx vy = vy0 -gt Δy = vy0t -½ gt 2 t = 3s [WROTE OUT THE Y DERIVATIONS ON THE LIGHT BOARD] By the way, this is not the ANSWER to the homework but this idea should look familiar if you ve done the homework already!

Generic Projectile Motion y motion: Same as vertical-only problem! x motion: Covers Δx distance at vx0 speed This is true for ALL PROJECTILES.

x and y: Motions are Independent Separation of vectors into components allows separations of equations into components: v x = Δx = v v xo xo + a t + x t 1 2 2 axt v y = v yo Δy = v yo + a t + y t 1 2 2 ayt ax = 0 m/s 2 ay = ±9.8 m/s 2 The main point is here that we can treat x and y motions independently. HERE IS WHERE THE DEMO WENT TERRIBLY WRONG :). The question we pose is: I am a monkey hunter wanting to hit the monkey hanging from the tree. As soon as I shoot, the monkey lets go of the tree Think about the x and y motion of the bullet and the monkey.

Monkey Hunter The instant the hunter fires, the monkey will reflexively let go of the tree and drop to the ground. Where should the hunter aim? A. Above the monkey. B. At the monkey. C. Below the monkey. Q24 A classic 2D physics problem.

Monkey Hunter If there were no gravity, the bullet would hit the monkey because the monkey would not move.

Monkey Hunter With gravity, the bullet and the monkey fall at the same rate: the rate of the acceleration of gravity, -9.8 m/s 2. The answer is the same no matter where the hunter is standing. (READ RED PART THEN SAY)- as long as the bullet has enough velocity to get to the monkey before it hits the ground. Another way of saying the same two things I ve been saying: 1. When solving kinematics problems, solve x and y piecewise. 2. Everything falls at the same rate.

Break up what you know in terms of the horizontal and vertical Prevents mistakes! X Y

After it leaves your hand, before it hits the ground or t vo = initial velocity vector θo = initial direction of velocity vector v y = 0 at top of trajectory v x = v xo remains the same throughout trajectory because there is no acceleration along the x-direction One other thing that s been confusing you: the initial and final point of these problems. You re often given a velocity and direction, and unless it notes that it s in the X or Y direction, that v will always be v0, the initial velocity vector. That theta will be the direction of the initial velocity vector. AS TIME CHANGES, THETA WILL CHANGE AS DEFINED BY ITS X AND Y COMPONENTS. You can see that here in this figure by the angle of the primary velocity vector arrow.

Strategy for Projectile Motion Problems Beyond this, projectile motion problems just take a lot of planning and thinking. Take your time and think about the set-up of the problem. What do I know? What s the first step? What s the next step? Extra important to follow our problem-solving tips: Break the problem down, Draw a picture, List knowns/unknowns.

Strategy for Projectile Motion Problems The time will be the same for x and y parts of the question. If you don t have enough information for x or y components, solve for time. Δy Δx Here s a far more specific and practical tip! Often times the first step is solving for time.

Throwing something off of a cliff (5 examples with increasing difficulty) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. 1. How much time does it take to fall? 2. How far from the base of the cliff does it hit the ground? (Need the time first) 3. How fast it is moving vertically when it hits the ground? (y component of final velocity)

How much time to fall? (Think Vertical) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. We re talking about something falling, and that is vertical motion, so we will only use vertical ideas and numbers. v = v 0 + at Δx = v 0 t + ½ at 2 v 2 = v 0 2 + 2aΔx Need to solve for time. Because x and y are independent, we ONLY CARE ABOUT Y HERE! That makes this problem easier.

How much time to fall? (Think Vertical) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. We re talking about something falling, and that is vertical motion, so we will only use vertical ideas and numbers. Δy = v oy t + 1/2 at 2 Δy = (16 sin 60) t +1/2 (-9.8) t 2 Quadratic eq. (at 2 +bt+c=0): a=-4.9, b=13.85, c=- Δy =15 m t=[-b ± (b 2-4(a)(c))^ ½] /(2a) =-13.85±(191.8-4(-4.9)(15))^ ½ ]/(-9.8) t= 1.41 ± 2.25= -0.84 s or 3.66 s [solution from light board above]

How far from cliff base? (Think Horizontal) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. We know it was in the air for 3.66s (from the previous question), and it s moving at a constant speed in the x-direction the whole time (a x = 0). v = v 0 + at Δx = v 0 t + ½ at 2 v 2 = v 0 2 + 2aΔx

How far from cliff base? (Think Horizontal) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. We know it was in the air for 3.66s (from the previous question), and it s moving at a constant speed in the x-direction the whole time (a x = 0). Δx = v ox t+ ½ a x t 2 where ax=0 Δx = v x t = (16 cos60º)(3.66s) [solution from light board above]

How far from cliff base? (Think Horizontal) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. We know it was in the air for 3.66s (from the previous question), and it s moving at a constant speed in the x-direction the whole time (a x = 0). Δx = v ox t+ ½ a x t 2 where ax=0 Δx = v x t = (16 cos60º)(3.66s) Δx = 29m [solution from light board above]

Vertical speed when it lands? (Think Vertical) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. Asking for y component of final velocity. It s been accelerating down the whole time. We know that gravity is causing this acceleration, so we can figure out how fast it is going (vertically) when it hits the ground. v = v 0 + at Δx = v 0 t + ½ at 2 v 2 = v 0 2 + 2aΔx

Vertical speed when it lands? (Think Vertical) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. Asking for y component of final velocity. It s been accelerating down the whole time. We know that gravity is causing this acceleration, so we can figure out how fast it is going (vertically) when it hits the ground. v fy = v iy + a t v fy = 16 sin60º -9.8 x3.66 v fy = -22 m/s Definitely negative is good, since moving opposite from positive y direction [solution from light board above]

Magnitude and Angle that it hits the ground? (Finally combine!) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. Final angle: which way is the ball going? How do we get is its final velocity vector? v = v 0 + at Δx = v 0 t + ½ at 2 v 2 = v 0 2 + 2aΔx The equations are here as a diversion.

Magnitude and Angle that it hits the ground? (Finally combine!) A ball is launched from the edge of a 15.0m tall cliff at 16 m/s at an angle of 60 degrees from the horizontal. c 2 = a 2 + b 2 = (8.0m/s) 2 + (-22m/s) 2 c = 23.4 m/s -22 tanθ = opp/adj = (-22 m/s) / (8m/s) Θ = tan -1 (-22/8) = -70 The object is moving at 23m/s at an angle of 70 below the horizontal when it hits the ground. [solution from light board above]