throwing at wall

D. Morin 3.47

You throw a ball with speed $v_0$ at a vertical wall, a distance $l$ away. At what angle should you throw the ball so that it hits the wall as high as possible? Assume $l < v_0^2/g$.

In general,

$$y = y_0 + v_{0,y}t - \frac{1}{2}gt^2 = y_0 + v_{0}\cdot\sin(\phi)\cdot t - \frac{1}{2}gt^2$$ $$x = x_0 + v_{0,x}t = x_0 + v_{0}\cdot\cos(\phi)\cdot t$$

The time it takes to hit the wall is,

$$t_{\text{wall}}=\frac{l}{v_0\cdot \cos(\phi)}$$

The height expression of the ball becomes,


In order to determine the angle that produces the maximum height when hitting wall, we need to take maximize this function with respect to the angle, $\frac{\partial h(\phi)}{\partial \phi}$.

Since, $$\frac{\partial}{\partial \phi}\tan(\phi) = \frac{\cos^2(\phi)+\sin^2(\phi)}{\cos^2(\phi)}=\sec^2(\phi)$$ $$\frac{\partial}{\partial \phi}\frac{1}{\cos(\phi)} = \frac{\sin(\phi)}{\cos^2(\phi)}=\sec(\phi)\cdot \tan(\phi)$$

$$\phi^* = \tan^{-1}\left( \frac{v_0^2}{gl} \right)$$