A ball having kinetic energy KE, is projected at an angle of 60° from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?

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$\frac{(KE)}{8}$

$\frac{(KE)}{4}$

$\frac{(KE)}{16}$

$\frac{(KE)}{2}$

answer is B.

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Detailed Solution

Initial K.E,
$K . E . = \frac{1}{2} {mu}^{2}$
Speed at highest point
$\begin{array}{l} V = ucos 60^{\circ} = \frac{u}{2} \\ ∴ {KE}_{2} = \frac{1}{2} m {(\frac{u}{2})}^{2} \\ = \frac{1}{4} \times \frac{1}{2} {mu}^{2} \\ = \frac{KE}{4} \end{array}$

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