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A particle having kinetic energy $K$ is projected at $60^{\circ}$ with the horizontal. The kinetic energy at the highest point is
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The correct answer is:
$K / 4$
The motion of projectile is as shown in figure,
Let initial velocity of projection is $u$, so initial kinetic energy is
$\mathrm{KE}_1=\frac{1}{2} m u^2$
At highest point, the vertical component of velocity is zero, while the horizontal component is $u \cos \theta$, which is non-zero. So, the kinetic energy at this point is
$\begin{aligned} \mathrm{KE}_2 & =\frac{1}{2} m(u \cos \theta)^2=\frac{1}{2} m u^2 \cos ^2 \theta \\ =\mathrm{KE}_1 \cos ^2 60^{\circ} & =\frac{K}{4}\left(\because \mathrm{KE}_1=K \text { and } \cos 60^{\circ}=1 / 2\right)\end{aligned}$
Let initial velocity of projection is $u$, so initial kinetic energy is
$\mathrm{KE}_1=\frac{1}{2} m u^2$
At highest point, the vertical component of velocity is zero, while the horizontal component is $u \cos \theta$, which is non-zero. So, the kinetic energy at this point is
$\begin{aligned} \mathrm{KE}_2 & =\frac{1}{2} m(u \cos \theta)^2=\frac{1}{2} m u^2 \cos ^2 \theta \\ =\mathrm{KE}_1 \cos ^2 60^{\circ} & =\frac{K}{4}\left(\because \mathrm{KE}_1=K \text { and } \cos 60^{\circ}=1 / 2\right)\end{aligned}$
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