Q52 / 75 · JEE Main 2025 (23 Jan Shift 2) · Physics

A ball having kinetic energy KE, is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ?

  • A
    $\frac{(\mathrm{KE})}{8}$
  • B
    $\frac{(\mathrm{KE})}{2}$
  • C
    $\frac{(\mathrm{KE})}{16}$
  • Correct answer
    D
    $\frac{(\mathrm{KE})}{4}$

Solution

Initial K.E, $\text { K.E. }=\frac{1}{2} \mathrm{mu}^2$ Speed at heighest point $\begin{aligned} & \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2} \\ & \therefore \mathrm{KE}_2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2 \\ & =\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^2 \\ & =\frac{\mathrm{KE}}{4} \end{aligned}$

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

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