Ball Thrown Over a Wall: Will It Clear, How Far Away, How High? (Projectile Motion at an Angle)
A ball is thrown at an angle toward a wall. Will it get over? How far away could the wall be and still be cleared? If it does not clear, how high up the wall does it hit? These three questions look different but are answered with the same method: resolve the velocity, find the time from the horizontal motion, then find the height from the vertical motion.
The problem
A student throws a ball from a height of 1.2 m above level ground with a speed of 22 m/s at 40° above the horizontal, toward a vertical wall 36 m away. Air resistance is negligible. Take g = 9.81 m/s². The ball just goes over the wall. Find the height of the wall. (An exam-style problem in our own words, not a copy of a paper.)
Step 1: choose signs and resolve the velocity
Take up as positive, so the vertical acceleration is −9.81 m/s² and the horizontal acceleration is 0.
Horizontal component: u_x = 22 cos 40° = 16.85 m/s. Vertical component: u_y = 22 sin 40° = 14.14 m/s. The vertical component is the one you will be asked for first, and it is the initial vertical velocity, not the speed 22 m/s.
Step 2: time to reach the wall (horizontal axis)
Horizontally the motion is uniform, so time = distance ÷ speed: t = 36 ÷ 16.85 = 2.136 s.
Step 3: height at the wall (vertical axis)
Use s = ut + ½at² vertically: s = 14.14 × 2.136 − ½ × 9.81 × 2.136² = 30.21 − 22.38 = 7.83 m above the launch point.
The ball started 1.2 m above the ground, so the height above the ground is 7.83 + 1.2 = 9.03 m. If the ball only just goes over the wall, the wall is 9.0 m high.
The rounding trap: 9.0 m or 9.3 m?
Round the time to 2.1 s and the same calculation gives 14.14 × 2.1 − 4.905 × 2.1² = 8.1 m, so 9.3 m. Using 2.136 s gives 9.0 m. The answers differ by 0.3 m because near the wall the ball is already falling at about 6.8 m/s, so a 0.036 s error in the time moves the height by about 0.25 m. Keep the unrounded time in your calculator and round only at the end.
Three ways the question can end
- •Will it clear the wall? Find the height of the ball at the wall (t = d ÷ u_x, then y = y₀ + u_y t − ½gt²) and compare it with the wall height. For this throw, a wall of 8 m is cleared by 1.03 m.
- •How far should the wall be so the ball only just clears it? Solve y = wall height for t. That is a quadratic, with two answers: the ball is at that height once going up and once coming down. The later time gives the farthest wall that is still just cleared. For an 8 m wall that is d = 38.3 m, and any wall from 10.3 m to 38.3 m away is cleared.
- •How high up the wall does it hit? If the wall is taller than the ball is at that distance, the ball hits at the height y you calculated. With a 12 m wall at 36 m, it hits 9.03 m up, 2.97 m below the top.
Quick sense checks
- •Highest point: y₀ + u_y² ÷ 2g = 1.2 + 14.14² ÷ 19.62 = 11.4 m. A wall taller than that can never be cleared at any distance.
- •Range: the ball comes back to the ground about 50.0 m from the thrower, so a wall farther than that is never reached.
- •The ball is at the top of its flight when its vertical velocity is zero, after u_y ÷ g = 1.44 s.
Frequently asked questions
Use the horizontal axis if you know the horizontal distance: t = d ÷ (u cos θ). If you need the time to land, solve the vertical equation y = 0, which is a quadratic in t.
Calculate the ball’s height at the wall’s horizontal distance. If that height is greater than or equal to the wall height, it clears; if it is less, it hits the wall at that height.
The ball rises through a given height on the way up and passes it again on the way down. Both give valid answers to the quadratic. A wall anywhere between those two positions is cleared.
Use the value your question or data booklet gives. The method is identical, but the numbers differ slightly, so do not mix values within one solution.