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The speed of a fall

Gravity and orbits · level 5, lesson 2 of 6

A question to start

A worker on a Mars building site slips off a 3 m platform. How fast do they hit the ground? Is it a lot gentler than the same slip on Earth, or only a little?

The idea

There are two ways to get the speed.

From the time: \(v = gt\).

From the height, without the time. From lesson 1, \(t = \sqrt{2s/g}\). Put that into \(v = gt\):

\[v = g\sqrt{\frac{2s}{g}} = \sqrt{2gs}, \qquad\text{or}\qquad v^2 = 2gs.\]

This is one of the most useful equations in mechanics. Speed grows with the square root of the height: four times the height gives only twice the landing speed.

Matching falls between planets. A fall from height \(s_M\) on Mars ends at the same speed as a fall from \(s_E\) on Earth when \(2g_M s_M = 2g_E s_E\), that is

\[s_E = s_M \times \frac{g_M}{g_E} = 0.38\, s_M.\]

A fall on Mars lands like a fall from 0.38 times the height on Earth.

Fall height Landing speed on Mars Landing speed on Earth
1 m 2.7 m/s 4.4 m/s
3 m 4.7 m/s 7.7 m/s
10 m 8.6 m/s 14.0 m/s

Worked examples

1. The platform. \(v = \sqrt{2 \times 3.71 \times 3} = \sqrt{22.3} = 4.7\) m/s. On Earth the same fall lands at \(\sqrt{2 \times 9.81 \times 3} = 7.7\) m/s. The Mars fall is like falling \(3 \times 0.38 = 1.1\) m on Earth: it can still hurt, but much less.

2. A ball thrown up. Thrown straight up at 12 m/s on Mars, how high does it go? At the top \(v = 0\), so run the equation backwards: \(12^2 = 2 \times 3.71 \times s\), so \(s = 144 / 7.42 = 19.4\) m. On Earth: \(144 / 19.62 = 7.3\) m.

3. Time to the top. \(t = v/g = 12 / 3.71 = 3.2\) s on Mars, and as long again to come back down.

Try it

  1. Find the landing speed after a 5 m fall on Mars.
  2. A rock lands at 10 m/s on Mars. From what height did it fall?
  3. What height on Earth gives the same landing speed as a 6 m fall on Mars?
  4. A ball is thrown up at 8 m/s on Earth. How high does it go? How long until it comes back?
  5. Show that a fall from four times the height gives twice the landing speed.
Answers
  1. \(\sqrt{2 \times 3.71 \times 5} = \sqrt{37.1} = 6.1\) m/s.
  2. \(s = v^2 / 2g = 100 / 7.42 = 13.5\) m.
  3. \(6 \times 0.378 = 2.3\) m.
  4. \(s = 64 / 19.62 = 3.3\) m; \(t = 2 \times 8 / 9.81 = 1.6\) s.
  5. \(\sqrt{2g \cdot 4s} = 2\sqrt{2gs}\).

Common mistake

Thinking the speed is proportional to height. Twice the height gives \(\sqrt2 \approx 1.41\) times the speed, not twice. It is the square of the speed that is proportional to the height.

Where it leads