WBJEE · Physics · Motion In Two Dimensions
A particle is projected at \(60^{\circ}\) to the horizontal with a kinetic energy \(\mathrm{K}\). The kinetic energy at the highest point is
- A K
- B zero
- C \(\frac{\mathrm{K}}{4}\)
- D \(\frac{\mathrm{K}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{K}}{4}\)
Step-by-step Solution
Detailed explanation
At highest point kinetic energy \(=1 / 2 \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=1 / 4 \times 1 / 2 \mathrm{~m} \mathrm{v} v^2=\mathrm{K} / 4\)
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