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A particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $\mathrm{K}$. The kinetic energy at the highest point is
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$\frac{\mathrm{K}}{4}$
At highest point kinetic energy $=1 / 2 \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=1 / 4 \times 1 / 2 \mathrm{~m} \mathrm{v} v^2=\mathrm{K} / 4$
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