A 180 pound person dives off of a 30 foot cliff. What is their velocity as they enter the water?

Provide step by step

Taking the sea level as reference for altitude, then, on top of the cliff the diver has only potential energy:

Ep=m×g×h, where m is the diver's mass, g=9.8m/s² and h=30ft=9.144m is the cliff's height.

At the sea level, the diver has kinetic energy: Ec=m×v²/2

From energy conservation: Ec=Ep, resulting v=sqrt(2×g×h)=13.4m/s

v=13.4m/s

Note that mass does not count in the final result.

that would be because of both gravity and friction. so     c