What if you threw a ball from a cliff?

Olly has a ball launcher. On flat ground, aiming halfway up sends the ball farthest.

What if the launcher sat high on a cliff? This is a pretend lab, so the cliff is only in the model.

Here is a famous rule of throwing: on flat ground, with no air to slow the ball, aiming at 45°, halfway between flat and straight up, sends it farthest.

But what if the ball starts high above the ground where it lands, like a ball launcher on top of a cliff? Does it still fly farthest at 45°? This lab answers that in a model. It is about the physics of flight only: in real life, stay well back from edges and never throw things from high places.

A projectile's horizontal and vertical motions are independent: OpenStax describes the horizontal velocity as constant while gravity changes the vertical one. From level ground that gives R = v₀² sin 2θ / g, maximized at θ = 45°.

The question here: if the launch point is a height h₀ above the landing ground, is 45° still optimal? This is a model of an idealized launcher on a 20 m cliff; it explores projectile physics and is never an activity to try near an edge.

On flat ground, Olly's launcher fires the ball at 45°.

The ball flies up, curves over and lands about 23 meters away.

Start on flat ground. The launcher fires the ball at 15 meters per second, aimed at 45°. With no air in the model, the ball flies along a curved path called a parabola and lands 22.9 meters away after 2.16 seconds, at its highest about 5.7 meters up.

Try other angles: any angle steeper or flatter than 45° lands shorter. Aimed at 30° or at 60°, the ball lands in the same spot, about 19.9 meters away.

At v₀ = 15 m/s, θ = 45°, h₀ = 0 and no drag, the model gives R = 22.9 m, a flight time of 2.16 s and a peak of 5.7 m, matching R = v₀² sin 2θ / g and T = 2v₀ sin θ / g.

Because sin 2θ is symmetric about 45°, complementary angles (30° and 60°, 15° and 75°) give equal ranges from level ground. That symmetry is what the cliff will break.

Now the launcher sits on top of a tall pretend cliff.

The ball has farther to fall. So it stays in the air longer. It lands much farther away!

Now raise the launch point to 20 meters, as if the launcher sat on a cliff above flat ground. Same speed, same 45° aim. The ball now flies for 3.37 seconds and lands 35.8 meters away.

Then try lower angles. At 31° the ball lands 38.0 meters out, the farthest of any angle in the model. From up high, a flatter launch wins. The ball also lands faster: 24.8 meters per second instead of 15.

With h₀ = 20 m, the 45° launch flies 3.37 s and lands at R = 35.8 m. Sweeping the angle in 1° steps, the range peaks at θ = 31°, R = 38.0 m (30° and 32° are within a few centimeters).

The landing speed is 24.8 m/s at every launch angle, since energy conservation fixes it: v² = v₀² + 2gh₀.

A steep throw spends its speed going up. A flatter throw spends more of it going forward.

From up high, the ball has lots of falling time anyway. So aiming flatter takes it farther.

A launch angle splits the ball's speed into two parts: up and forward. On flat ground, the ball needs upward speed to stay in the air long enough, and 45° is the perfect balance.

From a cliff, the ball gets extra time in the air for free while it falls the extra 20 meters. So it needs less upward speed, and putting more speed into going forward pays off. The higher the launch, the lower the best angle: in the model it is 40° from 5 meters, 36° from 10 meters and 31° from 20 meters.

The flight time T = (v_y + √(v_y² + 2gh₀))/g no longer falls to zero for a flat launch: with v_y = 0 it is still √(2h₀/g), about 2.0 s from 20 m. As h₀ grows, more of T comes from the drop itself, so trading v_y for vₓ = v₀ cos θ increases R = vₓT. The model's optimum falls from 45° (h₀ = 0) to 40° (5 m), 36° (10 m) and 31° (20 m).

Model limits: uniform gravity, flat landing ground, no spin or wind, and an optional quadratic drag with a constant drag coefficient. The ball stops where it first touches the ground.

Shot put athletes throw a heavy ball from up near their heads.

Because they let go up high, their best angle is lower than halfway up, too.

Athletes in the shot put push a heavy ball from about head height, so it lands below where it left their hand. The NRICH math project explains that this alone makes the best angle a little less than 45°.

Real people also can't push as hard at steep angles, which lowers the best angle more: in NRICH's worked solution, using a 2001 model by Linthorne published in the Journal of Sports Sciences, it came out around 35°. Air matters too. Switch it on in the model from the 20-meter cliff, and the best angle drops to about 27°.

NRICH sets the shot put as a raised-launch problem: released from around head height, the equal-speed optimum is slightly below 45°. Linthorne's 2001 model adds that release speed falls as the release angle rises, and NRICH's worked solution, for its chosen parameters, puts the optimum near 35°.

In our model, switching on quadratic drag (football-sized ball) from the 20 m cliff cuts the 31° range to about 30.5 m and moves the best angle to 27°. Real balls also spin and have drag coefficients that change with speed, which this smooth model leaves out.