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Need help with this physics problem please...?
A 68.2 kg diver steps off a 5 meter diving platform. Ignoring air resistance, what is the kinetic energy (KE) and the velocity of the diver as she enters the water?
I did the problem and I was wondering if I did it right....
m= 68.2 kg
height of the diving platform (hdp)= 5 meters
PEi= mgh= (68.2kg)(9.80m/s2)(5m)
= 3341.8 J and that would be the KE right?
For speed, KEf=1/2 *mvf^2
vf= square root of 2(KEf)/m which is square root of 2(3341.8J)/68.2kg
and that equals to 9.89949 m/s. Is this all correct?
Yes, correct.
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