27 feet = 8.23 m
PE =KE,
mgH = mv^2/2
v=sqroot(2gH)= 12.7 m/s.
If the collision with the ground is elastic, the height of the ball would be equal to its initial height. It cant reach the heigth of 15 m.
A ball is dropped from 27.0 feet above the ground. Using energy considerations only, what is the velocity of the ball prior to it hitting the ground? If the ball is thrown back up to a ledge 15m above the groud, what would be the velocity as it reaches the ledge?
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