Motion of a projectile
| English | Chinese | Pinyin |
|---|---|---|
| horizontal | 水平 | shuǐ píng |
| vertical | 竖直 | shù zhí |
| gravity | 重力 | zhòng lì |
| projectile | 抛射体 | pāo shè tǐ |
| parabola | 抛物线 | pāo wù xiàn |
| range | 射程 | shè chéng |
| greatest height | 最大高度 | zuì dà gāo dù |
The cannonball problem
- Fire a ball at an angle and it traces a smooth arch, then lands. Where, and how high?
- Galileo's insight: treat the horizontal 水平 and vertical 竖直 motions completely separately — they only share the clock.
Two independent motions
- Launched at speed $u$ and angle $\alpha$:
- Horizontal: constant velocity $u\cos\alpha$ (no force sideways).
- Vertical: gravity 重力 only, acceleration $g$ downward.
Water jets follow parabolic paths, the classic projectile 抛射体 motion under gravity
Launch a projectile
Fire the ball, then change the angle and speed. The horizontal motion is steady while gravity pulls it down — together they trace a parabola. Find the angle for the longest range, and try the Moon.
During the flight, the horizontal velocity of a projectile (ignoring air resistance):
There is no horizontal force, so u cos α is constant throughout.
The initial horizontal velocity of a projectile launched at speed u and angle α is u ______ α.
Horizontal component = u cos α; vertical = u sin α.
The parabola 抛物线
- Eliminating $t$ leaves $y$ as a quadratic in $x$ — so the path is a parabola.
Horizontal speed stays constant; gravity bends the vertical motion into a parabola.
The path of a projectile (no air resistance) is a:
Eliminating t gives y as a quadratic in x — a parabola.
Range 射程 and greatest height 最大高度
- At the top, the vertical velocity is momentarily zero. The key results:
Worked example. $u = 20\ \tfrac{\text{m}}{\text{s}}$, $\alpha = 30^\circ$, $g = 10$. Greatest height $= \dfrac{400 \times 0.25}{20} = 5\ \text{m}$.
A ball is thrown at u = 20 m/s at 30°. Using height = u²sin²α/(2g) with g = 10, find the greatest height (m).
400 × sin²30° / 20 = 400 × 0.25 / 20 = 5 m.
At what launch angle (degrees) is the range greatest?
Range ∝ sin 2α, which is largest when 2α = 90°, i.e. α = 45°.
At the highest point of the flight, the vertical component of velocity is zero.
Vertical velocity decreases to zero at the top, then becomes negative (downward).
Common trap
Range is maximised at $45^\circ$, not $90^\circ$. $\sin 2\alpha$ peaks when $2\alpha = 90^\circ$. Firing straight up gives zero range — all the speed goes into height.
- Eliminating time gives the Cartesian equation of the path; all trajectories lie under a bounding parabola.
You've got it
- resolve into horizontal ($u\cos\alpha$, constant) and vertical ($u\sin\alpha$, gravity) parts
- the path is a parabola
- range $= \dfrac{u^2\sin 2\alpha}{g}$ (max at $45^\circ$); greatest height $= \dfrac{u^2\sin^2\alpha}{2g}$