Mar4

Kinematic equations

optimizing take-off angle

Start with take-off velocity $v$ with components $$v = \begin{pmatrix}v_x \\ v_y\end{pmatrix}.$$ Assuming no air resistance, we know that air-time is $$\frac{2v_y}g.$$ Therefore, distance will be $$v_x\left(\frac{2v_y}g\right) = \frac{2v_xv_y}g.$$ Assuming some total speed $s$ and angle from horizontal $\theta$, we have $$\begin{align*} v_x = s\cos\theta, \\ v_y = s\sin\theta, \\ \end{align*}$$ so distance is $$\frac{2\cdot s\cos\theta\cdot s\sin\theta}g.$$

Since the double angle identity tells us that $\sin\theta\cos\theta = \sin(2\theta)/2$, this is equivalent to $$\frac{s^2\cdot\sin(2\theta)}{g}.$$

Since $\sin(2\theta)$ is maximized at $2\theta = \pi/2$ or $\theta = \pi/4$, we see that this distance is maximized for a given speed at the angle $\boxed{\theta = \pi/4.}$