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Question

A ball, whose kinetic energy is E, is projected at an angle of 45 from the horizontal. What will be it's kinetic energy at the highest point of its flight?

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Solution


Suppose the ball of mass m be projected with an initial velocity u.
E=12mu2
At the highest point, component of velocity in vertical direction vy=0
Component of velocity in horizontal direction, vx=ucos45=u2
So, kinetic energy of the ball at highest point,
K.E=12m(u2)2=(mu2)4=E2

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