Question

A 70 kg object strikes the ground with 2500 J of KE after falling freely from rest. How far above the ground was the object when it was released?

Answer 1

Answer:

3.64 m

Explanation:

m = Mass of object = 70 kg

Kinetic energy of the object = 2500 J

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

h = Height from which the object is dropped

Kinetic energy is given by

$\dfrac{1}{2}mv^2=2500\\\Rightarrow v^2=\dfrac{2\times 2500}{70}$

From conservation of energy we get kinetic energy equal to potential energy.

$\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{1}{2g}v^2\\\Rightarrow h=\dfrac{1}{2\times 9.81}\times \dfrac{2\times 2500}{70}\\\Rightarrow h=3.64\ \text{m}$

The object was released from a height of 3.64 m.