A ball is thrown upward with an initial velocity of 15 m/s at an angle of 60.0° above the horizontal. Use energy conservation to find the ball’s greatest height above the ground.
mvy2/2=mgh→h=v2sin2α/(2g)mv^2_y/2=mgh\to h=v^2\sin^2\alpha/(2g)mvy2/2=mgh→h=v2sin2α/(2g) . Answer
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