Projectile motionEdexcel International A Level Physics: Revision notes

A projectile is an object moving freely under gravity only, with air resistance negligible. Its motion splits into two independent parts:

• Horizontal: no force, so no acceleration and constant velocity, $s_x = v_x t$
• Vertical: constant acceleration $g = 9.81\ \text{m s}^{-2}$ downwards, so the equations of uniformly accelerated motion apply

Time is the only quantity shared by both parts. A ball dropped from rest and a ball projected horizontally from the same height reach the ground at the same time, because horizontal velocity does not change the vertical motion.

For an object launched horizontally at speed $u$ from height $h$:

• Vertical: $u_y = 0$, so $h = \frac{1}{2}gt^2$ and $t = \sqrt{\frac{2h}{g}}$
• Horizontal: range $= ut$
• At impact: $v_y = gt$ and the speed is $\sqrt{u^2 + v_y^2}$ at angle $\tan^{-1}(v_y/u)$ below the horizontal

Example: $12\ \text{m s}^{-1}$ from a $20\ \text{m}$ roof gives $t = \sqrt{40/9.81} = 2.0\ \text{s}$ and a range of $24\ \text{m}$.

For launch speed $u$ at angle $\theta$ above the horizontal, resolve the velocity:

• $u_x = u\cos\theta$ (constant)
• $u_y = u\sin\theta$ (changes with $a = -g$)

For level ground:

• At the highest point $v_y = 0$, so the maximum height is $\frac{u_y^2}{2g}$
• Time of flight $= \frac{2u_y}{g}$
• Range $= u_x \times$ time of flight

Example: $20\ \text{m s}^{-1}$ at $30°$ gives $u_x = 17\ \text{m s}^{-1}$, $u_y = 10\ \text{m s}^{-1}$, maximum height $5.1\ \text{m}$ and time of flight $2.0\ \text{s}$. At the top the velocity is $17\ \text{m s}^{-1}$ horizontally, not zero.

1. Resolve the initial velocity into horizontal and vertical components.
2. Choose a positive direction for the vertical motion (usually upwards, so $a = -9.81\ \text{m s}^{-2}$).
3. Use vertical suvat equations to find the time or the vertical position.
4. Use $s_x = u_x t$ for the horizontal distance.
5. To find the speed at any time, combine the components using Pythagoras.

Example: a ball kicked at $18\ \text{m s}^{-1}$ at $35°$ has $u_x = 14.7$ and $u_y = 10.3\ \text{m s}^{-1}$. After $20\ \text{m}$ horizontally, $t = 1.36\ \text{s}$ and the height is $10.3 \times 1.36 - \frac{1}{2} \times 9.81 \times 1.36^2 = 5.0\ \text{m}$.