Question

If a seagull drops a shell from rest at a height of 12 m how fast is the shell moving when it hits the rocks

Answer 1

As seagull drops a shell from rest at a height of 12 m, so we use kinematic equation of motion,

$v^{2} = u^{2} +2g h$

Here, h is the height, u is initial velocity , v is final velocity and g is acceleration due to gravity.

Given,  h = 12 m.

We take, $g = 9.8 \ m/s^2$ and $u = 0$ because seagull drops a  shell from rest.

Therefore, the speed of shell when it hits the rocks,

$v^{2} = 0 + 2 \times 9.8 m/s^2 \times 12 \ m = 235.2 (m/s)^2 \\\\ v = \sqrt{235.2 (m/s)^2} = 15.33 \ m/s^2$