As seagull drops a shell from rest at a height of 12 m, so we use kinematic equation of motion,
[tex]v^{2} = u^{2} +2g h[/tex]
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, [tex]g = 9.8 \ m/s^2[/tex] and [tex]u = 0[/tex] because seagull drops a shell from rest.
Therefore, the speed of shell when it hits the rocks,
[tex]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[/tex]
