INTRODUCTION. 3. Two-Dimensional Kinematics

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1 INTRODUCTION We now extend our study of kinematics to motion in two dimensions (x and y axes) This will help in the study of such phenomena as projectile motion Projectile motion is the study of objects that are initially launched (or projected) and then continue moving under the influence of gravity alone We will see that horizontal and vertical motions are independent For example a ball thrown horizontally with a speed v continues to move with the same speed v in the horizontal direction, even as it falls with an increasing speed in the vertical direction 1

2 MOTION IN 2D: CONSTANT VELOCITY Consider the situation shown above, where a turtle starts from the origin at t = 0, and moves with constant speed v 0 = 0.26m/s, 25 above the x axis In 5.0s, the turtle travels d = v 0 t = = 1.3m in a straight line In the x direction: x = d cos(25) = 1.2m In the y direction: y = d sin(25) = 0.55m Alternatively, we could treat the x and y motions separately The x velocity component is: v 0x = v 0 cos(25) = 0.24m/s The y velocity component is: v 0y = v 0 sin(25) = 0.11m/s Displacement x is: x = v 0x t = 1.2m Displacement y is: y = v 0y t = 0.55m Generally the turtle might start at a position x = x 0 at time t = 0 Thus: x = x 0 + v 0x t and y = y 0 + v 0y t 2

3 MOTION IN 2D: CONSTANT VELOCITY - EXAMPLE An eagle perched on a tree branch 19.5m above the water spots a fish swimming near the surface. The eagle pushes off from the branch and descend toward the water. By adjusting its body in flight, the eagle maintains a constant speed of 3.1m/s at an angle of 20 below the horizontal. How long does it take for the eagle to reach the water? How far has the eagle traveled in the horizontal direction when it reaches the water? 3

4 MOTION IN 2D: CONSTANT ACCELERATION In 1D, we used x = x 0 + v 0 t + ½ at 2 In 2D, we replace both v 0 and a with the corresponding x components, v 0x and a x Thus x = x 0 + v 0x t + ½ at 2 For the y direction: y = y 0 + v 0y t + ½ at 2 The above are position versus time equations of motion for two dimensions The same approach gives velocity as a function of time Start with v = v 0 + at and write it in terms of x and y components Thus: v x = v 0x + a x t and v y = v 0y + a y t We can write v 2 = v a x in terms of components Thus: v x2 = v 0x2 + 2a x x and v y2 = v 0y2 + 2a y y 4

5 MOTION IN 2D: CONSTANT ACCELERATION - EXAMPLE A hummingbird is flying in such a way that it is initially moving vertically with a speed of 4.6m/s and accelerating horizontally at 11m/s 2. Assuming the bird s acceleration remains constant for the time interval of interest, find the horizontal and vertical distance through which it moves in 0.55s. 5

6 PROJECTILE MOTION: BASIC EQUATIONS (1) Here we will consider the independence of horizontal and vertical motions to projectiles A projectile is an object that is thrown, kicked, batted, or otherwise launched into motion and then allowed to follow a path determined solely by gravity In studying projectile motion, the following assumptions are made Air resistance is ignored The acceleration due to gravity is constant, downward and has a magnitude g = 9.81m/s 2 The Earth s rotation is ignored Air resistance can be significant when a projectile moves with relatively high speed or in strong winds The value of g varies slightly from place to place on the Earth s surface and decreases with altitude The rotation of the Earth can be significant when considering projectiles that cover great distances Yet for the simple situation of dropping a ball, such drawbacks can be ignored 6

7 PROJECTILE MOTION: BASIC EQUATIONS (2) Consider the above figure, where the x axis is horizontal and the y axis is vertical, with upwards being the positive direction Since downwards is negative: a y = -9.81m/s 2 = -g Gravity causes no acceleration in the x direction, and so a x = 0 So for projectile motion we have the following: x = x 0 + v 0x t v x = v 0x v x2 = v 0x 2 y = y 0 + v 0y t - ½ gt 2 v y = v 0y - gt v y2 = v 0y2-2g y 7

8 PROJECTILE MOTION: BASIC EQUATIONS - DEMONSTRATION A simple demonstration illustrates the independence of horizontal and vertical motions in projectile motion Firstly while standing still, a rubber ball is dropped and caught on the rebound from the floor The ball goes straight down, lands near your feet and returns almost to the level of your hand Next walk or skate with constant speed and drop the ball, and observe its motion To you, the motion looks the same as when standing still, i.e. it lands near your feet and bounces straight back up The fact you were moving horizontally had no effect on the ball s vertical motion motions were independent To an observer who sees you walking by, the ball follows a curved path as shown 8

9 LAUNCH ANGLE OF A PROJECTILE A projectile launched at an angle above the horizontal where θ > 0 A launch below the horizontal would correspond to θ < 0 A projectile launched horizontally is when θ = 0 9

10 ZERO LAUNCH ANGLE: EQUATIONS OF MOTION A special case is a projectile is launched horizontally so that the angle between the initial velocity and the horizontal is θ = 0 Taking into consideration the figure on the previous slide, if we choose the ground level to be y = 0, with height h and walking speed v 0, we can say that the initial position of the ball is given by x 0 = 0 and y 0 = h The initial velocity is horizontal, which corresponds to θ = 0, and as a result the x and y velocity components are: v 0x = v 0 cos(0) = v 0 and v 0y = v 0 sin(0) = 0 Substituting these specific values into the fundamental equations for projectile motion yield: x = v 0 t v x = v 0 = constant v x2 = v 02 = constant y = h - ½ gt 2 v y = - gt v y2 = - 2g y Note that the x component of velocity remains the same for all time and that the y component steadily increases with time As a result, x increases linearly with time, and y decreases with t 2 dependence 10

11 ZERO LAUNCH ANGLE: EXAMPLE A person skateboarding with a constant speed of 1.3m/s releases a ball from a height of 1.25m above the ground. Given that x 0 = 0 and y 0 = h = 1.25m, find x and y for t = 0.25s and t = 0.5s. Find the velocity, speed and direction of motion of the ball at t = 0.5s. 11

12 PARABOLIC PATH Just what is the shape of the curved path followed by a projectile launched horizontally? This is found by combining x = v 0 t and y = h - ½ gt 2 Combining these equations allows us to express y in terms of x by eliminating t, i.e. t = x/v 0 Substituting this result into the y equation to eliminate t yields y = h ½ g(x/v 0 ) 2 = h (g/2v 02 )x 2 This has the form y = a + bx 2, where a = h and is constant Also b = g/2v 02 and is also constant This is the equation of a parabola that curves downwards, a characteristic shape in projectile motion 12

13 LANDING SITE Where does a projectile land if it is launched horizontally with a speed v 0 from a height h? The most direct way to solve this is to set y = 0, since y = 0 corresponds to ground level Thus 0 = h (g/2v 02 )x 2 Solving the above equation for x yields the landing site and thus x = v 0 (2h/g) Note that the positive sign for the square root has been chosen since the projectile was launched in the positive x direction, and hence lands at a positive value of x 13

14 PARABOLIC PATH AND LANDING SITE EXAMPLE A mountain climber encounters a crevasse in an ice field. The opposite size of the crevasse is 2.75m lower and is separated horizontally by a distance of 4.1m. To cross the crevasse, the climber gets a running start and jumps in the horizontal direction. What is the minimum speed needed by the climber to safely cross the distance? If, instead, the climber s speed is 6.0m/s, where does the climber land and what is the climber s speed on landing? 14

15 GENERAL LAUNCH ANGLE The more general case of a projectile being launched from an arbitrary angle with respect to the horizontal will be considered The simplifications made with zero launch angle are no longer used Here θ is nonzero For a projectile launched with initial speed v 0 at angle θ, the initial x and y positions are zero, as it starts at the origin, thus x 0 = y 0 = 0 The middle figure shows the components of the initial velocity The last figure shows the conditions when θ = 90 and 0 15

16 GENERAL LAUNCH ANGLE EQUATIONS Substituting the results from the previous slide yield the following x = (v 0 cosθ)t v x = v 0 cosθ v x 2 = v 0 2 cos 2 θ y = (v 0 sinθ)t ½ gt 2 v y = v 0 sinθ gt v y 2 = v 0 2 sin 2 θ 2g y These equations are valid for any launch angle They reduce to the simpler equations derived earlier when θ = 0 and y 0 = h Example: A projectile is launched from the origin with initial speed 20.0m/s at an angle of 35 above the horizontal. Find the x and y positions at times t = 0.5s, t = 1.0s and t = 1.5s. Also find the velocity at these times. 16

17 SNAPSHOTS OF A TRAJECTORY Above shows the projectile referred to in the previous example The points are not evenly spaced in terms of position even though they are spaced evenly in time The points bunch closer together at the top of the trajectory which indicates that a comparatively large fraction of the flight time is spent near the highest point This gives the illusion that an object hangs in the air, e.g. a basketball player performing a slam dunk 17

18 GENERAL LAUNCH ANGLE: EXAMPLE 1 Chipping from the rough, a golfer sends the ball over a 3.0m high tree that is 14.0m away. The ball lands at the same level from which it was struck after travelling a horizontal distance of 17.8m on the green. If the ball left the club 54 above the horizontal and landed on the green 2.24s later, what was the initial speed? How high was the ball when it passed over the tree? 18

19 GENERAL LAUNCH ANGLE: EXAMPLE 2 A golfer hits a ball from the origin with an initial speed of 30.0m/s at an angle of 50 above the level where the ball was struck. How long is the ball in the air? How far has the ball travelled in the horizontal direction when it lands? What is the speed and direction of motion of the ball just before it lands? 19

20 GENERAL LAUNCH ANGLE: EXAMPLE 3 A trained dolphin leaps from the water with an initial speed of 12.0m/s. It jumps directly toward a ball held by a trainer a horizontal distance of 5.5m away and a vertical distance of 4.1m above the water. In the absence of gravity the dolphin would move in a straight line to the ball and catch it, but because of gravity the dolphin follows a parabolic path well below the ball s initial position, as shown. If the trainer releases the ball the instant the dolphin leaves the water, show that the dolphin and the falling ball meet. 20

21 PROJECTILE MOTION: RANGE The range, R, is the horizontal distance a projectile travels before landing Consider the above where the initial and final elevations are the same (y = 0) To find the range use y = (v 0 sinθ)t ½ gt 2 to find t for y = 0 and substitute this time into the x equation of motion Thus for y = 0: (v 0 sinθ)t = ½ gt 2 so (v 0 sinθ) = ½ gt Therefore t = (2v 0 /g)sinθ (time of flight) Substitute this into x = (v 0 cosθ)t which yields x = (2v 02 /g)sinθcosθ, and x = R, sin2θ = 2sinθcosθ Thus the range R = (v 02 /g)sin2θ Above equation only valid for same initial and final elevation 21

22 RANGE: EXAMPLE A rugby game begins with a kickoff in which the ball travels a horizontal distance of 41m and lands on the ground. If the ball was kicked at an angle of 40 above the horizontal, what was its initial speed? Suppose the initial speed of the ball is increased by 10%. By what percentage does the range increase? 22

23 MAXIMUM RANGE R depends inversely on g, so the smaller g the larger the range On the moon, the acceleration due to gravity is about 1/6 of g, so a projectile would travel 6 times as far as it would on Earth What launch angle gives the greatest range? We know that R varies with sin2θ and so R is a maximum when sin2θ is a maximum, i.e. when sin2θ = 1 which is when sin(90) = 1, thus θ = 45 Since the maximum range occurs when sin2θ = 1 R max = (v 02 /g) only valid when the projectile lands at the same level from which it was launched Also it is only valid for the ideal case of no air resistance Air resistances come into consideration for fast moving objects, and the range is reduced The maximum range occurs for launch angles less that 45 23

24 SYMMETRY IN PROJECTILE MOTION (1) Recall that the time when the projectile lands is given by t = (2v 0 /g)sinθ By symmetry the time it takes a projectile to reach its highest point (in the absence of air resistance) should be half this time After all, the projectile moves in the x direction with constant speed, and the highest point (i.e. the maximum y) occurs at x = ½ R To prove this, it is safe to say that at the highest point the projectile is moving horizontally, so v y = 0 We would like to find the time when v y = 0 Thus v y = v oy gt = v 0 sinθ gt = 0 Therefore t = (v 0 /g)sinθ, which is half the time before the projectile lands 24

25 SYMMETRY IN PROJECTILE MOTION (2) There is another interesting symmetry for speed Recall that when a projectile is launched v y = v 0 sinθ When the projectile lands at t = (2v 0 /g)sinθ, v y is v y = v 0 sinθ gt = v 0 sinθ - g(2v 0 /g)sinθ = -v 0 sinθ This is the exact opposite of the v y when it was launched Since v x is always the same, it follows that when the projectile lands, its speed v = (v x2 + v y2 ) and is the same as when it was launched But the velocities are different since the directions of motion are different at launch and landing The symmetry described here extends to any level At a given height the speed of a projectile is the same on the way up as on the way down The angle of the velocity above the horizontal on the way up is same as the angle below the horizontal on the way down, as shown 25

26 SYMMETRY IN PROJECTILE MOTION (3) Consider the range R above left shows R versus launch angle θ for v 0 = 20m/s As discussed, without air resistance, R is a maximum when θ is 45 Note from the plot that the range for angles equally above or below 45 is the same, i.e. the range for 30 is the same for 60 26

27 MAXIMUM RANGE Recall that a projectile is at a maximum height when the y component of velocity is zero This can be used to determine the maximum height of an arbitrary projectile First, find the time when v y = 0 Then substitute this time into the y versus t equation where y = (v 0 sinθ)t ½ gt 2 Example: The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish s mouth. Suppose the archerfish squirts water with an initial speed of 2.3m/s at an angle of 19.5 above the horizontal. When the stream of water reaches a beetle on a leaf at height h above the water s surface, it is moving horizontally. How much time does the beetle have to react? What is the height h of the beetle? What is the horizontal distance d between the fish and the beetle when the water is launched? 27