All subjects · Mathematics · WME01 (M1)

Vertical motion under gravity

An object moving freely up or down has constant acceleration g = 9.8 m/s² downwards, so the suvat equations apply. The only new skill is handling signs.

Setting up

Choose a positive direction (usually upwards). Then a = -9.8 m/s². Displacements below the start are negative.

Model the object as a particle with no air resistance.

Key moments

At the highest point, v = 0.
It returns to its starting level with the same speed it was thrown at (but moving downwards).
"Lands 25 m below the point of projection" means s = -25 when up is positive.

Time in the air

Use s = ut + (1/2)at² with the final displacement, which often gives a quadratic. Reject the negative root. To find how long it is higher than some level, find both times when s equals that height and subtract.

Worked example

A ball is thrown vertically up at 14.7 m/s. Find its greatest height and the time to reach it.

Up positive: u = 14.7, v = 0, a = -9.8.
v² = u² + 2as: 0 = 216.09 - 19.6s, so s = 11.0 m.
v = u + at: 0 = 14.7 - 9.8t, so t = 1.5 s.

Answer: Greatest height 11.0 m, reached after 1.5 s

Key idea: Pick up as positive, so a = -9.8. Top: v = 0. Below the start: negative s. Reject negative times.

Check yourself

A ball is projected vertically upwards at 21 m/s. Find its greatest height, to 3 significant figures.
Show answer
22.5 m. 0 = 441 - 19.6s, so s = 22.5 m.
A stone is thrown upwards at 14.7 m/s. Find the total time until it returns to the point of projection.
Show answer
3 s. s = 0: 0 = 14.7t - 4.9t² = t(14.7 - 4.9t), so t = 3 s.
A stone is dropped from rest from a height of 40 m. Find the speed with which it hits the ground, to 3 significant figures.
Show answer
28.0 m/s. v² = 0 + 2(9.8)(40) = 784, v = 28 m/s.
A ball is thrown vertically upwards at 8 m/s from the top of a cliff 25 m above the sea. Find the time it takes to reach the sea, to 3 significant figures.
Show answer
3.22 s. Up positive: -25 = 8t - 4.9t², so 4.9t² - 8t - 25 = 0, giving t = 3.22 s (rejecting the negative root).
For the same ball (8 m/s up from a 25 m cliff), find its speed as it hits the sea, to 3 significant figures.
Show answer
23.5 m/s. v² = 8² + 2(-9.8)(-25) = 554, so v = 23.5 m/s.
A particle is projected upwards at 19.6 m/s. For how long is it at least 14.7 m higher than its starting point?
Show answer
2 s. 14.7 = 19.6t - 4.9t² gives t² - 4t + 3 = 0, so t = 1 or t = 3. It is at least 14.7 m higher for 3 - 1 = 2 s.
A ball is thrown vertically upwards. Which is true at its highest point?
- Its velocity and acceleration are zero
- Its acceleration is zero
- Its velocity is 9.8 m/s
- Its velocity is zero and its acceleration is 9.8 m/s² downwards
Show answer
Its velocity is zero and its acceleration is 9.8 m/s² downwards. Gravity still acts at the top, so the acceleration is still g downwards; only the velocity is momentarily zero.

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Last updated 7 November 2026. Spotted a mistake? Tell us.