Projectile Over a Fence: Which Launch Angle Works?
A ball is launched from ground level at 18 m/s toward a vertical fence 22 m away. The fence is 6.0 m high. The launch point is at ground level and air resistance is negligible. A student claims that launching at 45° will clear the fence because 45° gives maximum range. Another student says the ball cannot clear the fence at all because the maximum height at 45° is only about 8.3 m and the ball reaches the fence too late. The teacher asks for a quantitative judgement. Use g = 9.8 m/s².
1. Determine whether the ball launched at 45° clears the fence. Show the horizontal and vertical calculations separately.
Read the worked answer
Resolve the launch velocity: vx = 18 cos45° = 12.73 m/s; vy = 18 sin45° = 12.73 m/s.
Time to reach the fence (horizontal): t = x/vx = 22/12.73 = 1.728 s.
Height at that time: y = vy t − ½ g t² = 12.73(1.728) − 4.9(1.728)² = 22.0 − 14.63 = 7.37 m.
Since 7.37 m > 6.0 m, the ball clears the fence at 45° by about 1.4 m. The first student's conclusion is correct, but the reasoning ('45° gives maximum range') is not the reason-maximum range would put the ball at ground level at 33 m, and the relevant question is the height at x = 22 m, not the range. The second student's claim that the ball reaches the fence too late is also wrong: the ball is still rising or near its peak at t = 1.73 s (peak time is vy/g = 1.30 s), so it is descending but still well above 6 m.
2. Find the minimum launch angle (to the nearest degree) that just clears the fence at 18 m/s, and explain why there are two angles that clear it.
Read the worked answer
Let θ be the launch angle. Time to fence: t = 22/(18 cosθ). Height: y = 18 sinθ · t − 4.9 t². Setting y = 6.0 m and substituting:
6.0 = 22 tanθ − 4.9(22/(18 cosθ))² 6.0 = 22 tanθ − 4.9(484)/(324 cos²θ) 6.0 = 22 tanθ − 7.32(1 + tan²θ) [using 1/cos²θ = 1 + tan²θ]
Let u = tanθ: 7.32u² − 22u + 13.32 = 0. Solving: u = [22 ± √(484 − 390)]/14.64 = [22 ± 9.70]/14.64.
So u = 2.17 or u = 0.840, giving θ = 65.3° or θ = 40.0°.
The minimum angle is therefore about 40°. The two solutions arise because a steep launch gives a high, slow trajectory and a shallower launch gives a flatter, faster trajectory; both reach 6.0 m at x = 22 m. Angles below 40° give y < 6.0 m at the fence, and angles above 65° also fail because the ball falls short of the fence horizontally before descending to 6 m-the ball lands before reaching x = 22 m.