Question

A ball 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.

Answer 1

The answer is: 5.5

Please mark as brainliest!

Answer 2

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.