A projectile is fired from the top of a $40m$ high cliff with an initial speed of $50m/s$ at an unknown angle. Find its speed when it hits the ground.

Given , height $h=40m$

initial speed $u=50m/s$

Applying law of conservation of energy,

$mgh+\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2$

where, potential energy $U=mgh$

initial kinetic energy $K_i=\frac{1}{2}m{u}^2$

final kinetic energy $K_f=\frac{1}{2}m{v}^2$

$\Rightarrow gh+\frac{1}{2}{u}^2=\frac{1}{2}{v}^2$

$10\times 40+\frac{1}{2}\times {50}^2=\frac{1}{2}{v}^2$

$v^2=3300$

$\therefore v=57.4\approx 58m/s$