What if you dropped a ball from four times higher?

Olly drops a ball from a tall tower. It falls down to the ground.

Now he climbs four times higher. Will the ball take four times as long to land?

If you walk four times as far at the same speed, it takes four times as long. So if you drop a ball from four times as high, does it take four times as long to reach the ground?

It sounds sensible, but a falling ball does not move at a steady speed. In this lab you can drop a ball from 10 meters and from 40 meters in a model, time both falls, and see whether your guess holds up.

Distance equals speed times time only when speed is constant. A dropped object is different: OpenStax notes that, with air resistance and friction negligible, all objects at a given place fall with the same constant acceleration, independent of their mass. Its speed therefore grows the whole way down.

This lab asks how the fall time scales with the drop height. Using the fall model with air resistance off, compare h = 10 m with h = 40 m, then switch air on to see how drag changes the answer. Predict first: does the time double, quadruple or do something else?

Olly lets go of the ball. It starts slowly.

Then it goes faster and faster. It lands very quickly, in less than two seconds!

In the model, a ball dropped from 10 meters lands after about 1.4 seconds. It starts from rest, then speeds up by about 9.8 meters per second every second. By the time it hits the ground it is moving at about 14 meters per second.

Watch the dots the model leaves behind. They are a tenth of a second apart, and the gaps between them grow as the ball speeds up. Most of the distance is covered near the end of the fall.

With g = 9.81 m/s² and no air, the model gives a landing time of 1.43 s and a landing speed of 14.0 m/s for h = 10 m, matching t = √(2h/g) and v = √(2gh).

The distance fallen grows with the square of the time, h = ½gt². In the first half of the fall time the ball covers only a quarter of the drop, 2.5 m; the remaining 7.5 m pass in the second half. That uneven split is the key to the four-times question. The model steps the motion forward every 2 milliseconds and pins down the landing moment within the last step, so its times agree closely with these formulas.

Now Olly drops the ball from four times higher.

It does not take four times as long. It takes only twice as long! The ball is going very fast at the bottom.

Set the drop height to 40 meters. The ball falls four times as far, but it lands after about 2.9 seconds: only twice as long as from 10 meters.

Why not four times as long? Because the ball keeps gaining speed. During the extra time it is already moving fast, so it covers lots of ground quickly. It also lands twice as fast, at about 28 meters per second. To halve the fall time instead, you would need a drop only a quarter as high: 2.5 meters.

At h = 40 m the model gives 2.86 s and 28.0 m/s: both exactly double the 10 m values, since t ∝ √h and v ∝ √h. Four times the height gives √4 = 2 times the time; nine times the height would give three times the time.

At 2.5 m the time halves to 0.71 s. With air resistance switched on, the model's default football (0.45 kg, drag coefficient 0.47, 0.038 m²) takes 1.49 s from 10 m and 3.33 s from 40 m, a ratio of about 2.24, because drag, which grows with speed, matters more on the longer, faster fall.

A falling ball does not move at one speed. It speeds up all the way down.

So the second part of a long fall goes by very fast.

Gravity gives a falling ball the same push of extra speed every second. After one second it is moving at about 9.8 meters per second, after two seconds about 19.6. Because the speed keeps growing, the distance covered grows even faster: twice the time means four times the distance.

Turn that around and you get this lab's answer: four times the distance needs only twice the time. Our model is simple. It treats the ball as a dot, keeps gravity the same all the way down and leaves out air unless you switch it on.

Constant acceleration from rest gives v = gt and h = ½gt², so h ∝ t² and t ∝ √h. Equivalently, the average speed of the fall is half the landing speed, and the landing speed itself grows as √h, so the extra height is crossed at ever higher speed.

Model limits, from its documentation: uniform gravity over the drop, a rigid ball that does not spin, quadratic drag with a constant drag coefficient when air is on, and a ground that stops the ball with no bounce. The ball is drawn larger than to scale so it stays visible.

Long ago, a scientist named Galileo wanted to time falling balls. But falling is too quick!

So he rolled balls down a ramp instead. Rolling is slower and easier to time.

A University of Virginia physics lecture explains that the scientist Galileo Galilei wanted to know how speed changes during a fall, but a drop is over too quickly to measure. His trick was to roll a ball down a gentle ramp, which slows the motion down.

In his own account, the ball's distances, measured over and over, were always in proportion to the squares of the times. Twice the time meant four times the distance, just like the falling ball in our model. In real air, though, drag matters too: OpenStax explains that it grows with speed, so very long falls stop speeding up.

Michael Fowler's University of Virginia lecture on Galileo's Two New Sciences describes how Galileo argued that a falling body keeps picking up speed, then proposed uniform acceleration: equal gains of speed in equal times. To test it he rolled a ball down an inclined channel; in the experiment as Galileo reported it, repeated a full hundred times, the distances traversed were to each other as the squares of the times, for every inclination of the ramp.

That is exactly h ∝ t². Real falls through air depart from it at high speed: OpenStax shows drag growing until it equals the weight, when the object stops accelerating at its terminal speed. The model's air setting shows the start of that effect.