Clearing a Wall: Projectile Judgement
A football is kicked from level ground with an initial speed of 18 m/s at 35° above the horizontal. A vertical wall 2.5 m high stands 20 m from the kicker. The ball is modelled as a particle moving under gravity alone; air resistance is neglected and g = 9.81 m/s². A second barrier, 3.0 m high, is proposed for a point 25 m from the kicker, and the coach wants to know whether the same kick would clear it. You should resolve the initial velocity into horizontal and vertical components, use the horizontal motion to fix the time at each barrier, and then judge the height. The coach is particularly interested in whether the ball is rising or falling as it passes the first wall, because that affects how much clearance a small change in angle would give.
1. Determine whether the ball clears the 2.5 m wall at 20 m, and state whether it is rising or falling at that point.
Read the worked answer
Resolve the initial velocity: horizontal component u_x = 18 cos 35° = 14.745 m/s; vertical component u_y = 18 sin 35° = 10.325 m/s.
At the first wall x = 20 m, the time is t = x / u_x = 20 / 14.745 = 1.356 s. The height is y = u_y t − ½gt² = 10.325 × 1.356 − 4.905 × (1.356)² = 13.999 − 9.019 = 4.98 m.
This is above 2.5 m, so the ball clears the wall by about 2.48 m. The time to maximum height is u_y / g = 10.325 / 9.81 = 1.052 s. Since 1.356 s is later than 1.052 s, the ball is falling as it passes the wall. That matters because on the descending part of the path the clearance decreases with each extra metre, so a small reduction in angle or speed could turn a comfortable clearance into a failure.
2. Find the maximum height and the horizontal range of the ball, and use them to decide whether it clears the proposed 3.0 m barrier at 25 m and where it lands relative to a fence at 32 m.
Read the worked answer
Maximum height: y_max = u_y² / (2g) = 10.325² / 19.62 = 5.43 m. Range: R = u² sin 2θ / g = 18² × sin 70° / 9.81 = 324 × 0.9397 / 9.81 = 31.04 m. So the ball lands about 31.0 m from the kicker, before a fence at 32 m.
At x = 25 m, t = 25 / 14.745 = 1.695 s. Then y = 10.325 × 1.695 − 4.905 × (1.695)² = 17.50 − 14.09 = 3.41 m. This is above the proposed 3.0 m barrier by about 0.41 m, so the same kick would clear it, but only narrowly. Because the ball is descending, the clearance is sensitive to the assumed launch speed and angle, so the coach should not treat 0.41 m as a large safety margin.