Projectiles

The idea

Split the motion into horizontal and vertical parts. Horizontally there is no acceleration: $x = (u\cos\theta)t$ . Vertically the acceleration is $g$ downwards, so use SUVAT with $u\sin\theta$ . Time links the two.

Resolve the initial velocity: $u\cos\theta$ across, $u\sin\theta$ up.
Use vertical SUVAT to find the time (to the top, or to land).
Use that time horizontally for the distance.

Watch out Only the vertical motion has an acceleration; the horizontal speed stays the same.

See it worked

Worked example

A ball is projected at $20\text{ m s}^{-1}$ at $30^\circ$ above horizontal ground. Find its range.

Components: $20\cos 30^\circ = 17.32$ across, $20\sin 30^\circ = 10$ up.
Vertical: $0 = 10t - 4.9t^2$ , so $t = \frac{10}{4.9} = 2.041\text{ s}$ .
Range $= 17.32 \times 2.041 = 35.3\text{ m}$ .

Answer $35.3\text{ m}$

Your turn

Question 1Calculator allowed3 marks

A stone is thrown horizontally at $15\text{ m s}^{-1}$ from a cliff $20\text{ m}$ high. Find the time it takes to hit the sea. Use $g = 9.8\text{ m s}^{-2}$ and give your answer to 3 significant figures.

Question 2Calculator allowed3 marks

For the same stone, find how far from the foot of the cliff it lands, to 3 significant figures.

Question 3Calculator allowed3 marks

A ball is projected at $25\text{ m s}^{-1}$ at $40^\circ$ above the horizontal. Find its greatest height above the point of projection, to 3 significant figures.

In past papers

29 official questions in 29 papers since 2018. The latest:

All A-level Maths past papers

A-level Maths Pro, Pearson Edexcel 9MA0

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