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

The correct option is C $58\text{m/s}$
Given data, height of the cliff, $h=40\text{m}$
Initial speed of the projectile, $u=50\text{m/s}$
Let the projectile hit the ground with velocity ${}^{\prime }{v}^{\prime }$
Applying the law of conservation of energy,
$mgh+\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2$
$10\times 40+\frac{1}{2}\times 2500=\frac{1}{2}{v}^2$
$v^2=3300$
$v=57.4\text{m/s}=58\text{m/s}$