What if you tossed a ball straight up?

Throw a ball up high. Up it goes, then down it comes!

Is it going fast or slow when it comes back down? Let's find out.

Everyone has tossed a ball straight up and watched it come back. It rises, seems to hang in the air for a moment, then drops back into your hands.

But what is really happening to its speed on the way up and on the way down? Is it slower when it comes back, faster, or the same? In this lab you can toss a ball in a model and read its speed at every moment.

A vertical toss is the simplest case of free fall with an initial velocity. OpenStax's physics text stresses that free fall does not only mean falling down: once the ball leaves your hand, the same equations describe the way up and the way down.

The question is about symmetry. Neglecting air, how does the time up compare with the time down, and how does the speed on the way back compare with the launch speed? Then: what changes when air pushes back?

First, Olly just lets go of the ball. He does not throw it.

The ball falls down. It goes faster and faster until it lands.

Start with the simplest case. The ball starts 2 meters above the ground and Olly simply lets go. Gravity pulls it down, and it gains about 9.8 meters per second of speed every second.

In the model, with no air in the way, the dropped ball lands after 0.64 seconds, moving at 6.3 meters per second. Watch the speed readout climb the whole way down.

Released from rest at h = 2 m with g = 9.81 m/s² and no drag, the ball lands at t = √(2h/g) = 0.64 s with speed √(2gh) = 6.3 m/s. The velocity grows linearly with time, v = −g·t, because the only force is the ball's weight.

This is our ordinary case: zero launch velocity, one-way motion, speed rising from 0 to its landing value.

Now Olly tosses the ball up, fast!

It goes up and slows down. At the very top it stops for a tiny moment. Then it falls back down.

Now Olly tosses the ball straight up at 8 meters per second. Watch the speed readout: it drops as the ball climbs, reaches zero at the top, then rises again as the ball falls.

In the model the ball climbs from 2 meters to about 5.3 meters in 0.82 seconds. It takes another 0.82 seconds to fall back to 2 meters, and at that moment it is moving at 8.0 meters per second, exactly its starting speed. It then keeps falling and lands after 1.85 seconds at 10.2 meters per second.

With v₀ = 8 m/s upward, the model gives a rise time v₀/g = 0.82 s and a peak of h + v₀²/2g = 5.3 m. The ball passes 2 m again at 2v₀/g = 1.63 s with speed 8.0 m/s, and lands at 1.85 s at 10.2 m/s, the extra speed coming from the last 2 m of fall.

The speed graph is a V: a straight line down to zero, then a straight line back up with the same slope. Its slope is g, which never changes, even at the top where the velocity is momentarily zero.

Gravity pulls on the ball all the time, going up and coming down.

On the way up, the pull slows it. On the way down, the pull speeds it up by the same amount.

Gravity never switches off, not even at the top. On the way up it takes away about 9.8 meters per second of speed every second. On the way down it gives back exactly the same amount every second.

So the ball loses its speed going up and gains it all back coming down. With no air to steal any of it, the ball returns to the starting height just as fast as it left. The model only follows the ball's center: it leaves out spin and wind.

The acceleration is −g throughout, so v(t) = v₀ − g·t. The ball is at its starting height when v₀t − ½gt² = 0, at t = 2v₀/g, where v = −v₀: same speed, opposite direction. Energy gives the same answer: with no drag, kinetic plus gravitational potential energy is constant, so equal heights mean equal speeds.

The model assumes uniform g and treats the ball as a point with no spin; its drag option is quadratic, F = ½ρC_dAv², with a constant drag coefficient. Drag breaks the symmetry because it always opposes the motion and removes energy on both legs.

Real air pushes on a ball a little.

So a real ball comes back a bit slower than it went up. Catch it gently!

OpenStax's physics textbook works through a real example: a baseball hit straight up and caught 5.0 seconds later. Ignoring air, it left the bat at 24.5 meters per second, rose for 2.5 seconds to about 30.6 meters, fell for 2.5 seconds, and came back at 24.5 meters per second.

Real air pushes back harder the faster a ball moves. Switch air resistance on in the model: the same 8 meter-per-second toss now only climbs to about 5.0 meters and comes back past its starting height at about 7.4 meters per second.

OpenStax's worked example of a baseball hit straight up and caught after 5.0 s gives v₀ = 24.5 m/s, a 30.6 m peak at 2.5 s and a return speed of 24.5 m/s, and notes that a ball returns with the speed it had when it left as a general property of free fall.

Air breaks that rule. For baseballs and other large objects at everyday speeds, drag grows roughly with the square of speed. With the model's drag switched on (football-sized ball, C_d = 0.47), the 8 m/s toss peaks near 5.0 m and passes 2 m on the way down at about 7.4 m/s instead of 8.0. Spin, wind and the ball's changing drag coefficient are not modeled.