Question 51 of 183
A ball having kinetic energy KE, is projected at an angle of from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ?
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Answer: (D)
Initial K.E,
Speed at heighest point
$\begin{aligned}
& \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2} \\
& \therefore \mathrm{KE}_2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2 \\
& =\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^2 \\
& =\frac{\mathrm{KE}}{4}
\end{aligned}$
Speed at heighest point
$\begin{aligned}
& \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2} \\
& \therefore \mathrm{KE}_2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2 \\
& =\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^2 \\
& =\frac{\mathrm{KE}}{4}
\end{aligned}$