Question

A stone is gently dropped from a height of 80 m. If its velocity increases uniformly at the rate of 10 ms^-2, with what velocity will it strike the ground?

Answer 1

Given:

The height (h) from where the stone is dropped = 80 m.

The acceleration due to gravity (g) of the stone = 10 ms^-2.

Required to find:

The velocity (v) of the stone when it strikes the ground.

Solution:

The final velocity (v) of a freely falling object is given by $v=\sqrt{2gh}$.

By substituting the given values in the above equation, we get

$v=\sqrt{2x10x80}=\sqrt{1600}=\sqrt{{\left(40\right)}^2}=40m{s}^{-1}$

Hence, the velocity of the stone when it strikes the ground is 40 ms^-1.