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A lead shot is projected from the ground level with a velocity U, at an angle O, to the horizontal. ...

A lead shot is projected from the ground level with a velocity U, at an angle O, to the horizontal. Given that the times t, for the lead shot to reach its maximum height as t=UsinO/g. Where g is acceleration due to gravity. Show that H reached by the body is Hmax=U^2sin^2O/2g

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