ExamBro
ExamBro
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.
Verified Solution

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
\(\begin{aligned}
& \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2} \\
& \therefore \mathrm{KE}_2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2 \\
& =\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^2 \\
& =\frac{\mathrm{KE}}{4}
\end{aligned}\)
Same subject
Explore more questions on app