Projectile over a Practice Wall
During a training session, a player kicks a ball from level ground with initial speed 20 m/s at 35° above the horizontal, aiming at a vertical practice wall 25 m away. The wall is 8 m high, and the ball must pass over it to reach the target area behind. The coach uses g = 9.8 m/s² and ignores air resistance. An assistant says the ball will clear the wall because the maximum height of the trajectory is about 6.7 m; the coach points out that maximum height occurs before or after the wall, so the height at the wall itself must be checked. After the first kick fails, the player wants to know the minimum speed, keeping the same angle, that would just clear the wall. The ball is treated as a particle, and the wall is thin enough that its thickness can be ignored.
1. Determine whether the ball clears the 8 m wall. Compute the horizontal and vertical motion at the wall, and comment on the assistant's maximum-height argument.
Read the worked answer
Resolve the initial velocity: ux = 20 cos 35° ≈ 16.38 m/s and uy = 20 sin 35° ≈ 11.47 m/s. The time to reach the wall horizontally is t = 25/16.38 ≈ 1.526 s. The vertical displacement at that time is y = uy t − 0.5 g t^2 ≈ 11.47(1.526) − 0.5(9.8)(1.526)^2 ≈ 17.51 − 11.41 = 6.10 m. Since 6.10 m < 8 m, the ball does not clear the wall. The assistant's maximum-height argument is invalid because the maximum height is ymax = uy^2/(2g) ≈ (11.47)^2/19.6 ≈ 6.71 m and occurs at t = uy/g ≈ 1.17 s, which corresponds to a horizontal distance of about 19.2 m, before the wall. After that point the ball is descending, so its height at the wall is less than the maximum.
2. Find the minimum initial speed at the same angle that makes the ball just pass over the wall. Show the equation and the numerical result.
Read the worked answer
Use the trajectory equation y = x tan θ − g x^2/(2u^2 cos^2 θ), with θ = 35°, x = 25 m and y = 8 m. Then 8 = 25 tan 35° − 9.8(25)^2/(2u^2 cos^2 35°). Since 25 tan 35° ≈ 17.505, rearranging gives 9.8(625)/(2u^2 cos^2 35°) ≈ 17.505 − 8 = 9.505. Thus u^2 ≈ 9.8(625)/(2(0.6710)(9.505)) ≈ 6125/12.755 ≈ 480.2, so u ≈ 21.91 m/s. The minimum speed is therefore about 21.9 m/s, keeping the same angle. At this speed the ball passes exactly through the top of the wall; at the original 20 m/s it falls short, consistent with the first calculation.