*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!

**Best answer:**

*Answer by billrussell42*

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!**

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)