## College Physics (4th Edition)

(a) The maximum range is $\frac{v_i^2}{g}$ (b) The maximum range occurs at a launch angle of $45^{\circ}$
$R = \frac{2~v_i^2~sin~\theta~cos~\theta}{g} = \frac{v_i^2~sin~2\theta}{g}$ (a) The maximum range occurs when $sin~2\theta = 1$. Then the maximum range is $\frac{v_i^2}{g}$ (b) The maximum range occurs when $sin~2\theta = 1$. We can find the angle $\theta$: $sin~2\theta = 1$ $2\theta = sin^{-1}~(1)$ $2\theta = 90^{\circ}$ $\theta = 45^{\circ}$ The maximum range occurs at a launch angle of $45^{\circ}$