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A particle is projected at an angle $30^{\circ}$ with horizontal having kinetic energy $K$. The kinetic energy of the particle at highest point is.
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Verified Answer
The correct answer is:
$\frac{3}{4} K$
Initial kinetic energy is
Final kinetic energy is
$\begin{array}{rlrl}
K^{\prime} & =\frac{1}{2} m v^2=\frac{1}{2} m\left(u \cos 30^{\circ}\right)^2 & {\left[\because v=u \cos 30^{\circ}\right]} \\
& =\frac{3}{4}\left(\frac{1}{2} m u^2\right)=\frac{3}{4} K & & \text { [From Eq. (i)] }\end{array}$
Final kinetic energy is
$\begin{array}{rlrl}
K^{\prime} & =\frac{1}{2} m v^2=\frac{1}{2} m\left(u \cos 30^{\circ}\right)^2 & {\left[\because v=u \cos 30^{\circ}\right]} \\
& =\frac{3}{4}\left(\frac{1}{2} m u^2\right)=\frac{3}{4} K & & \text { [From Eq. (i)] }\end{array}$
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