TS EAMCET · Physics · Motion In Two Dimensions
A particle is projected with velocity \(2 \sqrt{g h}\) and at an angle \(60^{\circ}\) to the horizontal so that it just clears two walls of equal height \(h\) which are at a distance \(2 h\) from each other. The time taken by the particle to travel between these two walls is
- A \(2 \sqrt{\frac{2 h}{g}}\)
- B \(\sqrt{\frac{h}{2 g}}\)
- C \(2 \sqrt{\frac{h}{g}}\)
- D \(\sqrt{\frac{h}{g}}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{\frac{h}{g}}\)
Step-by-step Solution
Detailed explanation
Given, velocity of particle \(=2 \sqrt{g h}\) Angle \((\theta)=60^{\circ}\) Distance covered by particle \(=2 h\) Now, \[ 2 h=2 \sqrt{g h} \cos 60^{\circ} t \] or, \(\sqrt{h}=\sqrt{g} \times \frac{1}{2} t\) or, \(t=2 \sqrt{\frac{h}{g}}\)
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