Chapter 12.6, Problem 65ES

A shell is fired from ground level at an elevation angle of $\alpha$ and a muzzle speed ${\upsilon }_0.$

(a) Show that the maximum height reached by the shell is maximum height $\frac{{\left({\upsilon }_0\sin \alpha \right)}^2}{2g}$

(b) The horizontal range R of the shell is the horizontal distance traveled when the shell returns to ground level. Show that $R=\raisebox{1ex}{$\left({\upsilon }_0^2\sin 2\alpha \right)$}\!\left/ \!\raisebox{-1ex}{$g$}\right..$ For what elevation angle will the range be maximum? What is the maximum range?