A 180-pound person dives off of a 30-foot cliff. What is his velocity as he enters the water?
13.4 m/s.
588 m/s.
178 m/s.
24 m/s.
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
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