Question
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.

Answer

h = 40m, u = 50m/sec Let the speed be ‘v’ when it strikes the ground. Applying law of conservation of energy,$\text{mgh}+\frac{1}{2}\text{mu}^2+\frac{1}{2}\text{mv}^2$
$\Rightarrow10\times40+\frac{1}{2}\times2500=\frac{1}{2}\text{v}^2$
$\Rightarrow\text{v}^2=3300$
$\Rightarrow\text{v}=57.4\text{m}/\text{sec}\approx58\text{m}/\text{sec}$

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