At what angle should a ball be thrown with a speed of $56{\mathrm{ms}}^{-1}$ such that it covers the maximum possible range without hitting the 40 m high ceiling of the hall?

Here, $\mathrm{u}=56{\mathrm{ms}}^{-1}$

Let $\theta$ be the angle of projection with the horizontal to have maximum range, with maximum height = 40 m

Maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$

$\begin{array}{l}40=\frac{(50{)}^2{\sin }^2\theta }{2\times 9.8}\\ {\sin }^2\theta =\frac{2\times 9.8\times 40}{(56{)}^2}=\frac{1}{4}\text{ or }\sin \theta =\frac{1}{2}\text{ or }\theta \\ ={\sin }^{-1}\left(\frac{1}{2}\right)={30}^{\circ }\end{array}$