A 180-pound person dives off of a 30-foot cliff. What is his velocity as he enters the water?

A 180-pound person dives off of a 30-foot cliff. What is his velocity as he enters the water?

Detailed Explanation

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, \(g=9.8m/s^²\) and \(h=30ft=9.144m\) is the cliff's height.

At the sea level, the diver has kinetic energy: \(Ec=m×v^²/2\)

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

Answer: v=13.4m/s

Ep=m×g×h, where m is the diver's mass, \(g=9.8m/s^²\) and \(h=30ft=9.144m\) is the cliff's height.

At the sea level, the diver has kinetic energy: \(Ec=m×v^²/2\)

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

Answer: v=13.4m/s