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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 23 January · Evening Shift · Physics · Work, Power and Energy

A ball having kinetic energy KE, is projected at an angle of 60^...

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

Options

  1. A(KE)8\frac{(\mathrm{KE})}{8}
  2. B(KE)2\frac{(\mathrm{KE})}{2}
  3. C(KE)16\frac{(\mathrm{KE})}{16}
  4. D(KE)4\frac{(\mathrm{KE})}{4}

Correct answer

D. (KE)4\frac{(\mathrm{KE})}{4}

Step-by-step solution

Initial K.E,  K.E. =12mu2Speedatheighestpoint\text{ K.E. }=\frac{1}{2} \mathrm{mu}^2 Speed at heighest pointV=ucos⁡60∘=u2∴KE2=12m(u2)2=14×12mu2=KE4\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}$$
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