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Q.
A particle having kinetic energy $K$ is projected at $60^{\circ}$ with the horizontal. The kinetic energy at the highest point is
TS EAMCET 2020
Solution:
The motion of projectile is as shown in figure,
Let initial velocity of projection is $u$,
so initial kinetic energy is
$K E_{1}=\frac{1}{2} m u^{2} \ldots$ (i)
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
$K E_{2}=\frac{1}{2} m(u \cos \theta)^{2}=\frac{1}{2} m u^{2} \cos ^{2} \theta$
$=K E_{1} \cos ^{2} 60^{\circ}=\frac{K}{4}$
$\left(\because K E_{1}=K\right.$ and $\left.\cos 60^{\circ}=1 / 2\right)$
