Answer to Question #160622 in Mechanics | Relativity for Quanta

Question

A rock is thrown upward from level ground in such a way that the maximum height of its flight is equal to its horizontal range R. At what angle u is the rock thrown?

Answer 1

The vertical and horizontal components are, respectively, $v\sin \alpha, v\cos\alpha$. The time to the highest point (where the vertical component vanishes) is $T=\frac{v\sin \alpha}{g}$, and by symmetry the total time until the rock falls is $t=2T = \frac{2v\sin\alpha}{g}$ . The maximum height attained can be expressed as $h=\frac{v^2\sin^2\alpha}{2g}$ (for a uniformly accelerated movement we have $d= \frac{v_{final}^2-v_{initial}^2}{2a}$, where $v_{final},v_{initial}, a$ are taken in the direction of movement). The horizontal range can be expressed as $R=v\cos \alpha \cdot t=\frac{2v^2\sin\alpha \cos\alpha}{g}$, as in the horizontal direction the movement is uniform (we neglect the air resistance etc). Therefore, $R=h$ is achieved when $2\sin\alpha \cos \alpha=\frac{\sin^2\alpha}{2}$. This equation has two solutions (we take $0\leq \alpha \leq \pi/2$) : $\alpha=0$ (the obvious one, if the rock is thrown parallel to the ground from the level of the ground, it does not travel at all) and $\tan \alpha = 4$, $\alpha=\arctan 4 \approx 76^\circ$.