Launched at an Angle split a launch into horizontal motion at constant velocity and vertical motion with $a = -g$. Calculus adds two powers: setting a derivative to zero finds the top of the path, and Optimization finds the angle that sends a ball farthest. Try it in the Trebuchet lab after this lesson.
With launch speed $v_0$ at angle $\theta$ from level ground (no air resistance): $$x(t) = (v_0\cos\theta)t, \qquad y(t) = (v_0\sin\theta)t - \tfrac{1}{2}gt^2.$$
Key idea
The highest point is where $\dfrac{dy}{dt} = v_y = 0$: at $t = \dfrac{v_0\sin\theta}{g}$, with height $H = \dfrac{(v_0\sin\theta)^2}{2g}$. The range on level ground is $R = \dfrac{v_0^2\sin 2\theta}{g}$, largest at $\theta = 45^\circ$. Launched from a height, the best angle is less than $45^\circ$.
Why it works
The range comes from landing when $y = 0$ again: $t\left(v_0\sin\theta - \tfrac{1}{2}gt\right) = 0$ gives $t = 2v_0\sin\theta/g$, and $R = v_0\cos\theta \cdot 2v_0\sin\theta/g = v_0^2\sin 2\theta/g$. To maximize, differentiate with respect to the angle: $\dfrac{dR}{d\theta} = \dfrac{2v_0^2\cos 2\theta}{g} = 0$ when $2\theta = 90^\circ$. The second derivative is negative there, so it's a maximum. Steeper throws stay up longer but move sideways slowly; flatter throws move sideways fast but land quickly. At $45^\circ$ the trade is balanced.
Worked example: Maximum height
A ball is launched at $20\ \text{m/s}$ at $60^\circ$. Find its maximum height and its speed there.
- $v_y = 20\sin 60^\circ - 9.8t = 17.32 - 9.8t = 0$ at $t = 1.77\ \text{s}$.
- $H = \dfrac{(17.32)^2}{2(9.8)} = 15.3\ \text{m}$.
- At the top only $v_y$ is zero; $v_x = 20\cos 60^\circ = 10\ \text{m/s}$, so the speed there is $10\ \text{m/s}$.
Answer: $H \approx 15.3\ \text{m}$; speed at the top $10\ \text{m/s}$.
Worked example: Throwing from a cliff
From a height $h$, the range as a function of angle is messier, but the result of setting $dR/d\theta = 0$ is $\tan\theta^* = \dfrac{v_0}{\sqrt{v_0^2 + 2gh}}$. Find $\theta^*$ for $v_0 = 15\ \text{m/s}$ and $h = 20\ \text{m}$.
- $\sqrt{v_0^2 + 2gh} = \sqrt{225 + 392} = 24.84\ \text{m/s}$ (this is the landing speed).
- $\tan\theta^* = 15/24.84 = 0.604$, so $\theta^* = 31.1^\circ$.
- The extra fall time rewards a flatter, faster-moving throw.
Answer: $\theta^* \approx 31.1^\circ$.
Common mistake
Setting the whole velocity to zero at the top. If the ball stopped at the top it would fall straight down, and the path would come to a point instead of a smooth arch. Only $dy/dt$ is zero there; $dx/dt$ never changes. Check: the speed at the top equals $v_0\cos\theta$, the smallest speed of the flight but not zero (unless the throw was straight up).
Air drag lowers the best angle on level ground below $45^\circ$ too, often to 35°–40° for a ball. The equations here leave drag out; the Quadratic Drag topic shows how big it is.