JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2025
- Easy
- MCQ
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 ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{(\mathrm{KE})}{4}\)
Step-by-step Solution
Detailed explanation
Initial K.E,
\(\text { K.E. }=\frac{1}{2} \mathrm{mu}^2\)
Speed at heighest point
\(\begin{aligned}
& \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2} \\
& \therefore \mathrm{KE}_2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2 \\
& =\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^2 \\
& =\frac{\mathrm{KE}}{4}
\end{aligned}\)
\(\text { K.E. }=\frac{1}{2} \mathrm{mu}^2\)
Speed at heighest point
\(\begin{aligned}
& \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2} \\
& \therefore \mathrm{KE}_2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2 \\
& =\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^2 \\
& =\frac{\mathrm{KE}}{4}
\end{aligned}\)
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