AP EAMCET · PHYSICS · Motion In Two Dimensions
A body is projected at an angle of \(60^{\circ}\) with the horizontal. If the initial kinetic energy of the body is X, then its kinetic energy at the highest point is
- A X
- B 2X
- C \(\frac{X}{2}\)
- D \(\frac{X}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{X}{4}\)
Step-by-step Solution
Detailed explanation
Initial KE: \(K_i = X = \frac{1}{2}mv^2\) Velocity at highest point: \(v_h = v \cos \theta\) KE at highest point: \(K_h = \frac{1}{2}mv_h^2 = \frac{1}{2}m(v \cos \theta)^2 = \frac{1}{2}mv^2 \cos^2 \theta\) \(K_h = X \cos^2 60^{\circ}\)…
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