3.4.1.4
Horizontal projection: initial vertical velocity is zero
Projectile Motion — AQA A-Level Physics
- If an object is projected horizontally (e.g. rolling off a table), $u_{y} = 0$.
- Use $s = 0.5gt^2$ to find the time to fall height h.
- Then use horizontal $distance = u_{x} times t$ to find how far it lands.
Worked Example
A motorcycle takes off horizontally from a height of 1.25 m and lands 10 m away. What was the take-off speed?
Show Solution
1
Find the fall time from vertical motion
$$s = \frac{1}{2}gt^2$$
$$1.25 = \frac{1}{2} \times 9.81 \times t^2$$
$$t = \sqrt{\frac{2 \times 1.25}{9.81}} = 0.505 \text{ s}$$2
Find the horizontal speed
$$u_x = \frac{\text{horizontal distance}}{t} = \frac{10}{0.505} = 19.8 \text{ m s}^{-1} \approx 20 \text{ m s}^{-1}$$
Answer
Take-off speed $\approx 20$ m s$^{-1}$
Related:Kinematics