# Use energy conservation to find the ball’s greatest height above the ground?

Question by paperhearts63: Use energy conservation to find the ball’s greatest height above the ground?
A ball is thrown upward with an initial velocity of 16.0 m/s at an angle of 50.0 degrees above the horizontal. Use energy conservation to find the ball’s greatest height above the ground. I’m sure I’m doing this right but it told me my answer is wrong. Can someone explain how to do this? Thanks!

Getting the velocity component that is vertical, that is 16sin50 = 12.3 m/s.

Kinetic Energy in J if m is in kg and V is in m/s
KE = ½mV²
Potential Energy in J
PE = mgh
g is the acceleration of gravity 9.8 m/s²

set KE equal to PE, and solve for h
h = v²/2g = (12.3)²/2(9.8) = 7.66 meters

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What do you think? Answer below!

## One thought on “Use energy conservation to find the ball’s greatest height above the ground?”

1. Melanie Brashow says:

M x G x H = M x V^2 /2
(G = 9,8; V is velocity: 16.0 m/s; V^2 = V x V)
The angle does not matter.
Thus, G x H = V x V /2
H = V x V / (2 x G)