A kicked ball, a thrown javelin, a stone from a catapult: each starts with a velocity at an angle. Split it into components, and you have a vertical launch and a steady horizontal motion running side by side, exactly as for a horizontal launch.
Key idea
Launch at speed $v_0$ and angle $\theta$: $v_{0x} = v_0\cos\theta$ stays the same all flight; $v_{0y} = v_0\sin\theta$ changes by $-9.8\ \text{m/s}$ each second. On level ground the path is symmetric: up time equals down time.
Why it works
Only gravity acts during the flight, and it is vertical. So the vertical motion is a throw straight up with speed $v_{0y}$, and the horizontal motion is steady at $v_{0x}$. At the top only $v_y$ is zero; the ball still moves sideways at $v_{0x}$. For level ground the range is $v_{0x} \times \dfrac{2v_{0y}}{g} = \dfrac{2v_0^2\sin\theta\cos\theta}{g}$, which is the same for $\theta$ and $90^\circ - \theta$.
Worked examples
Worked example: Time, height and range
A ball is kicked at $20\ \text{m/s}$ at $30^\circ$ above level ground.
- Components: $v_{0x} = 20\cos 30^\circ \approx 17.3\ \text{m/s}$; $v_{0y} = 20\sin 30^\circ = 10\ \text{m/s}$.
- Time to the top: $\dfrac{10}{9.8} \approx 1.02\ \text{s}$, so the flight lasts $\approx 2.04\ \text{s}$.
- Height: $\dfrac{v_{0y}^2}{2g} = \dfrac{100}{19.6} \approx 5.10\ \text{m}$.
- Range: $17.3 \times 2.04 \approx 35.3\ \text{m}$.
Answer: $2.04\ \text{s}$, $5.10\ \text{m}$ high, $35.3\ \text{m}$ away
Worked example: Velocity at the top
For the same kick, what is the ball's velocity at the top?
- $v_y = 0$ there, and $v_x$ never changed.
- So $\vec{v} = 17.3\ \text{m/s}$, horizontal.
Answer: $17.3\ \text{m/s}$, horizontal
Worked example: Landing lower than the launch
From a wall $10\ \text{m}$ high, a ball is thrown at $15\ \text{m/s}$, $40^\circ$ above horizontal. When does it hit the ground?
- $v_{0y} = 15\sin 40^\circ \approx 9.64\ \text{m/s}$; the ground is at $\Delta y = -10\ \text{m}$.
- $-10 = 9.64t - 4.9t^2$, so $4.9t^2 - 9.64t - 10 = 0$.
- Positive root: $$t = \dfrac{9.64 + \sqrt{9.64^2 + 196}}{9.8} \approx 2.72\ \text{s}.$$
Answer: about $2.72\ \text{s}$
Common mistake
Saying the velocity is zero at the top of the path. Only the vertical part is zero; the horizontal part is the same as at launch. (Only a ball thrown straight up stops at the top.) Check: at the top, speed $= v_0\cos\theta$, the smallest speed of the flight but not zero.
Try it in the Science Lab's Trebuchet Workshop: change the release angle and see the range. With air drag, the best angle drops a little below $45^\circ$.