WBJEE · Physics · Motion In Two Dimensions
A particle is projected from the ground with a kinetic energy \(E\) at an angle of \(60^{\circ}\) with the horizontal. Its kinetic energy at the highest point of its motion will be
- A \(E / \sqrt{2}\)
- B \(E / 2\)
- C \(E / 4\)
- D \(E / 8\)
Answer & Solution
Correct Answer
(C) \(E / 4\)
Step-by-step Solution
Detailed explanation
At ground, \(E=\frac{1}{2} m u^{2}\) At highest point, \(E^{\prime}=\frac{1}{2} m\left(u \cos 60^{\circ}\right)^{2}\) \[ E^{\prime}=\frac{E}{4} \]
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