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 * 9.8 m/s^2 * 12 \ m = 235.2 (m/s)^2 \\\\ v = √(235.2 (m/s)^2) = 15.33 \ m/s^2`