a shot is fired at an angle of 60 to the horizontal with KE energy E.if air resistance is ignored the KE at the top of the tajectory is
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Let initial velocity be u.
Then initial horizontal velocity = u cos60° = u/2
Thus E = 1/2 m u^2
At the top velocity of the shot = u/2
Thus K.E. = 1/2 m (u/2)^2 = E/4
Then initial horizontal velocity = u cos60° = u/2
Thus E = 1/2 m u^2
At the top velocity of the shot = u/2
Thus K.E. = 1/2 m (u/2)^2 = E/4
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