A boy stands $78.4 m$ away from a building and throws a ball which just enters a window at maximum height $39.2m$ above the ground. Calculate the velocity of projection of the ball.

 

Maximum height $= \frac{u^2 {\sin }^2\theta }{2g} = 39.2 m \dots \left(i\right)$

Range $= \frac{u^2 \sin 2\theta }{g} = \frac{2u^2 \sin \theta \cos \theta }{g} = 2 \times 78.4 \dots \left(ii\right)$

from equation $\left(i\right)$ divided by equation $\left(ii\right)$   $\tan \theta = 1 \Rightarrow \theta = {45}^{\circ }$

from equation $\left(ii\right)$ range $=\frac{u^2 \sin {90}^{\circ }}{g} = 2\times 78.4 \Rightarrow u = \sqrt{2\times 78.4\times 9.8} = 39.2 m/s$