Q139 / 160 · WBJEE 2012 · Physics

A particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be

  • A
    $E / \sqrt{2}$
  • B
    $E / 2$
  • Correct answer
    C
    $E / 4$
  • D
    $E / 8$

Solution

At ground, $E=\frac{1}{2} m u^{2}$ At highest point, $E^{\prime}=\frac{1}{2} m\left(u \cos 60^{\circ}\right)^{2}$ $$ E^{\prime}=\frac{E}{4} $$

Hence, the correct answer is option: 'C'.

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