Motion of a projectile
Projectile motion
- A projectile moves under gravity: acceleration $g$ down, no horizontal acceleration.
- Treat horizontal and vertical separately. Launched at speed $u$, angle $\alpha$:
$$x = u\cos\alpha\,t, \qquad y = u\sin\alpha\,t - \tfrac12 g t^2$$
- The path is a parabola. Range $= \dfrac{u^2\sin 2\alpha}{g}$; greatest height $= \dfrac{u^2\sin^2\alpha}{2g}$.
Practice
A ball is thrown at u = 20 m/s at 30°. Using height = u²sin²α/(2g) with g = 10, what is the greatest height (m)?
height = 20² × sin²30° / (2×10) = 400 × 0.25 / 20 = 100/20 = 5 m.
Practice
In projectile motion, the horizontal acceleration is:
There is no horizontal force, so horizontal acceleration is zero; only the vertical part has g.
Practice
The trajectory of a projectile is a parabola.
Eliminating t from the equations gives a parabolic Cartesian path.
You've got it
Key idea
- resolve into horizontal ($u\cos\alpha$, steady) and vertical ($u\sin\alpha$, gravity) parts
- the trajectory is a parabola
- range $= \dfrac{u^2\sin 2\alpha}{g}$, max height $= \dfrac{u^2\sin^2\alpha}{2g}$