Projectile Over a Wall: Angle Versus Speed
A football is kicked from level ground at 20 m/s at an angle of 30° above the horizontal, directly toward a vertical wall 25 m away. The wall is 5.0 m high and has negligible thickness. Take g = 10 m/s² and ignore air resistance. The kicker wants the ball to clear the wall and land at least 3.0 m beyond it, so the landing point must be at horizontal distance 28 m or more from the kick-off point. A teammate claims that simply kicking harder at the same angle is the only fix; another suggests changing the angle while keeping the speed at 20 m/s. The ball is treated as a point particle, and the ground is horizontal on both sides of the wall.
1. At the stated 30° launch, determine whether the ball clears the 5.0 m wall. Show the horizontal and vertical calculations and state the ball's height when it reaches the wall.
Read the worked answer
Horizontal velocity: u_x = 20 cos 30° = 17.32 m/s. Time to reach the wall: t = 25 / 17.32 = 1.443 s. Vertical velocity: u_y = 20 sin 30° = 10.0 m/s. Height at the wall: y = u_y t − ½ g t² = 10.0(1.443) − 5(1.443)² = 14.43 − 10.41 = 4.02 m. Since 4.02 m < 5.0 m, the ball reaches the wall below its top edge; it does not clear the wall. The teammate's original kick fails, so either speed or angle must change.
2. Keeping speed at 20 m/s, find the minimum launch angle above 30° that just clears the wall. Explain why a second, larger angle also gives a trajectory through the top of the wall, and why it may not satisfy the landing requirement.
Read the worked answer
At speed 20 m/s, the trajectory must pass through (x, y) = (25, 5). Using y = x tanθ − gx²/(2u²cos²θ), with cos²θ = 1/(1+tan²θ): 5 = 25 tanθ − 10(25)²/(2×20²) = 25 tanθ − 7.8125(1+tan²θ). Let T = tanθ: 7.8125T² − 25T + 12.8125 = 0. Discriminant = 625 − 400.39 = 224.61; √ = 14.99. T = (25 ± 14.99)/15.625 gives T = 2.559 or 0.640, so θ = 68.7° or 32.6°. Both trajectories pass through the wall top: one is a high lob, one is a flatter drive. The smallest angle above 30° is about 32.6°. Check landing: for 32.6°, R = u² sin2θ/g = 400 sin65.2°/10 = 36.3 m, which is at least 28 m. For 68.7°, R = 400 sin137.4°/10 = 27.1 m, only 2.1 m beyond the wall, so it fails the required 3.0 m clearance. Thus 32.6° is the suitable angle.
3. If the kicker insists on 30°, what minimum speed is needed to clear the wall, and does that speed also satisfy the 28 m landing requirement?
Read the worked answer
At θ = 30°, require y = 5 at x = 25: 5 = 25 tan30° − 10(25)²/(2u²cos²30°) = 14.434 − 4166.7/u². Therefore 4166.7/u² = 9.434, u² = 441.7, u = 21.0 m/s. Range at this speed: R = u² sin60°/g = 441.7(0.866)/10 = 38.3 m. This exceeds 28 m, so the ball clears the wall and lands more than 3 m beyond it. The minimum speed is about 21 m/s; the teammate's 'kick harder' idea can work, but it requires about 5% more speed than 20 m/s, whereas changing the angle to 32.6° achieves the same clearance at the original speed. So the angle change is the more efficient fix.