First slide

Rectilinear Motion

Question

A body is projected with kinetic energy K at an angle of 60° with the horizontal. Its kinetic energy at the highest point of its trajectory will be

Moderate

A

2 K
B

K
C

K/2
D

K/4

Solution

At the highest point, the velocity has only the horizontal component $v_{x} = v \cos θ = v \cos 60^{\circ} = v / 2 .$
Now kinetic energy $\frac{1}{2} {mv}^{2}$ is proportional to v². Since the velocity is reduced to half, the kinetic energy becomes one-fourth, i.e., K/4.

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