Question

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

Answer 1

Answer:

The projectile hits the ground with a speed of 57.3 m/s.

Explanation:

Height of the cliff, h = 40 m

Initial speed of the projectile, u = 50 m/s

Let the projectile hit the ground with final velocity 'v'.

By applying the law of conservation of energy,

$PE{initial+KE_{initial}=PE_{final}+KE_{final}}\\mgh+\frac{1}{2} mu^2=mgh'+\frac{1}{2} mv^2\\m\times 9.8\times 40+\frac{1}{2} m\times50^2=m\times 9.8\times 0+\frac{1}{2} mv^2\\392\times 2+1250\times 2=v^2\\v=\sqrt{3284} \\v=57.3 m/s$

The projectile hits the ground with a speed of 57.3 m/s.