Question #253949

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.


1
Expert's answer
2021-10-28T08:54:31-0400

mvy2/2=mghh=v2sin2α/(2g)mv^2_y/2=mgh\to h=v^2\sin^2\alpha/(2g) . Answer


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