# physics

a stone is thrown straight up from a 50m high building and hits the ground at a speed of 37.1m/s. What is the initial velocity and what is the maximum height above the ground that the stone reaches?

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1. For maximum height, equate PE and KE
mgh=(1/2)mv²

h=(1/2)37.1²/9.8=70.2 m

at 50m, KE is reduced by mgh
=>
(1/2)mv²=(1/2)m(37.1)²-mg(50)
v=sqrt((37.1²-2(g)(50))
=19.9 m/s

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