Initial K.E, K.E. =21mu2 Speed at heighest point $\begin{aligned}
& \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2} \
& \therefore \mathrm{KE}_2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2 \
& =\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^2 \
& =\frac{\mathrm{KE}}{4}
\end{aligned}$
A ball having kinetic energy KE, is projected at an angle of 60° from the horizontal. What will be the kinetic energy of ball at the highest point of…
JEE Main 2025 — Physics Mechanics
A ball having kinetic energy KE, is projected at an angle of 60∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ?
Held on 23 Jan 2025 · Verified 6 Jul 2026.
Options
- A
8(KE)
- B
2(KE)
- C
16(KE)
- D
4(KE)
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