Aball is released from the top of a hill. how fast is the ball going when it reaches the base of the hill? approximate g as 10 m/s2 and round the answer to the nearest tenth. h1= 2m h2=0.5m no other information is given.

First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.

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