Projectile Motion

1. Projectile Motion For a rifle with a given muzzle velocity and an angle of elevation of 20°: Because vR is at an angle above the horizontal, it can be resolved into a horizontal and vertical velocity. vR vR = 75 m/s θ θ= 20°

2. vR vv θ vH vR is the resultant velocity, vv is the verticalvelocity, and vH is the horizontal velocity. Assuming no air resistance, the free-body diagram of the bullet after firing the gun is given by: FW

3. R Because Fnet = 0 in the horizontal direction, the bullet will try to travel in a straight line at constant speed in accordance with Newton’s 1st Law. The path of the projectile is called the trajectory and the path of the bullet will be parabolic. vR y vv θ θ vH

4. R is called the range of the projectile and y is the maximum height. The horizontal velocity is constant in the absence of air resistance. • vH = Δx/Δt = R/Δt = vRcosθ • Δt = hang time (twice the free fall time)

5. Vertically, there is deceleration and acceleration. • vf2 = vi2 + 2gΔy • vf = vi + gΔt • y = viΔt + 1/2gΔt2 • g = 9.80 m/s2 = 980 cm/s2 = 32 ft/s2

6. R Projectile Problem A boy throws a baseball with a velocity of 5.67 m/s at an angle of 56° above the horizontal. (a) How long is the ball in the air? vR y vv θ vH

7. vR = 5.67 m/s θ = 56° vv = vRsinθ = 5.67 m/s × 0.829 = 4.70 m/s vf = Vi + gΔt 0 = 4.70 m/s – 9.80 m/s2Δt Δt = 0.480 s ΔtT = 2 × Δt = 2 × 0.480 s = 0.960 s

8. (b) What is the maximum height the ball rises? vf2 = vi2 + 2gΔy 0 = (4.70 m/s)2 + 2 × -9.80 m/s2 × Δy y = 1.13 m

9. (c) Determine the range of the ball? vH = Δx/Δt Δx = vRcosθΔt Δx = 5.67 m/s × 0.853 × 0.960 s = 4.64 m (d) What angle would give the maximum range? 45° always gives the maximum range for a given velocity.

10. (e) What other angle would give the same range? Any two complementary angles give the same range, therefore 34° will also give the same range. For the larger angle: • vH is smaller, vV is larger, and the hang time is greater.

11. For the smaller angle: • vH is larger, vV is smaller, and the hang time is shorter.

12. vv vH 3.0 m 40. m Another Projectile Problem The game is on the line. There are two seconds left to play and the ball has been spotted 40. m from the uprights. If the ball is kicked with an initial speed of 30. m/s at an angle of 22° with the horizontal and the horizontal bar on the uprights is 3.0 m high, is the field goal kicker a hero or a goat? vR θ

13. Δx = 40. m vR = 30. m/s g = 9.80m/s2 Δy = 3.0 m θ = 22° vv = vR × sinθ = 30. m/s × sin22° = 11.2 m/s vH = vR × cosθ = 30. m/s × cos22° = 27.8 m/s vf = vi + gΔt Δt = 11.2 m/s/9.80 m/s2 = 1.14 s (1) Δt = 2 × 1.14 s = 2.28 s

14. vH = Δx/Δt Δt = 40. m/27.8 m/s = 1.4 s (2) Δy = y0 + viΔt + 1/2gΔt2 Δy = 0 + 0 + 1/2 × 9.80 m/s2 × (1.4 s – 1.14 s)2 Δy = 0.33 m (3)

15. vf2 = vi2 + 2gΔy 0 = (11.2 m/s)2 – 2 × 9.80 m/s2 × Δy Δy = 6.40 m (4) Δy = 6.40 m – 0.33 m = 6.07 m (5) Δy = 6.07 m – 3.0 m = 3.1 m (6) The field goal kicker gets to live another game by clearing the goal post by 3.1 m.

16. Some Final Thoughts There is a shorter way to do this problem but it is nowhere near as instructive. Each numbered step will be explained. (1) The time needed to reach its maximum height will be used to determine if the football is ascending or descending.

17. (2) Because the time needed to travel 40. m (1.4 s) is greater than the time needed to reach the highest point (1.14 s), the football must be descending. (3) How far has the ball fallen from its highest point in 1.4 s – 1.14 s = 0.26 s? (4)What is the maximum height of the football? (5)When the football reaches the goalpost, how high it is above the ground?

18. (6) Because the height of the football is greater than the height of the goalpost, subtracting the two heights gives how much the football clears the goalpost.