JEE Mains · Physics · STD 11 - 3.2 motion in plane
A ball is projected with kinetic energy \(E\), at an angle of \(60^{\circ}\) to the horizontal. The kinetic energy of this ball at the highest point of its flight will become.
- A \(Zero\)
- B \(\frac{E}{2}\)
- C \(\frac{E}{4}\)
- D \(E\)
Answer & Solution
Correct Answer
(C) \(\frac{E}{4}\)
Step-by-step Solution
Detailed explanation
\(E =\frac{1}{2} mu ^{2}\) At Highest point, Velocity \(V = u \cos 60^{\circ}=\frac{ u }{2}\) \(\therefore K E \text { at topmost point }=\frac{1}{2} m \left(\frac{ u }{2}\right)^{2}=\frac{E}{4}\)
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