Question 236 of 308

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 :

AZeroB$\dfrac{E}{2}$C$\dfrac{E}{4}$D$E$

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Answer: (C) $\dfrac{E}{4}$

$E=\dfrac{1}{2}m{u}^{2}$
At the Highest point, velocity is equal to the horizontal component of the velocity of projection. Therefore,
$v=u\cos 60^{\circ}=\dfrac{u}{2}$
$∴$ Kinetic energy at the topmost point $=\dfrac{1}{2}m{\left(\dfrac{u}{2}\right)}^{2}=\dfrac{E}{4}$