A ball is projected with kinetic energy E, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :
Solution
<p>$K.E. = E = {1 \over 2}m{v^2}$</p>
<p>at highest point</p>
<p>$K.E' = {1 \over 2}m{v^2}{\cos ^2}\theta$</p>
<p>$= {1 \over 2}m{v^2}\left( {{1 \over 4}} \right)$</p>
<p>$= {E \over 4}$</p>
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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