TS EAMCET · Physics · Motion In Two Dimensions
A projectile object is thrown in the upward direction making an angle of \(60^{\circ}\) with the horizontal with velocity of \(140 \mathrm{~m} / \mathrm{s}\). Then, the time after which its velocity makes an angle \(30^{\circ}\) with the horizontal is (Take, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(\frac{14}{\sqrt{3}} S\)
- B \(7 \sqrt{3} \mathrm{~s}\)
- C \(14 \sqrt{3} \mathrm{~s}\)
- D \(\frac{7}{\sqrt{3}} \mathrm{~s}\)
Answer & Solution
Correct Answer
(A) \(\frac{14}{\sqrt{3}} S\)
Step-by-step Solution
Detailed explanation
Velocity of the projectile, \(v=140 \mathrm{~m} / \mathrm{s}\) The angle of projection, \(\theta=60^{\circ}\). The horizontal component of the velocity is always constant throughout the motion. Let the velocity is \(u\) when the angle is \(30^{\circ}\) with horizontal.…
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