AP EAMCET · PHYSICS · Motion In Two Dimensions
If the range of a body projected with a velocity of \(60 \mathrm{~m} \mathrm{~s}^{-1}\) is \(180 \sqrt{3} \mathrm{~m}\), then the angle of projection of the body is
(Acceleration due to gravity \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) )
- A \(30^{\circ}\) or \(60^{\circ}\)
- B \(37^{\circ}\) or \(53^{\circ}\)
- C \(20^{\circ}\) or \(70^{\circ}\)
- D \(15^{\circ}\) or \(75^{\circ}\)
Answer & Solution
Correct Answer
(A) \(30^{\circ}\) or \(60^{\circ}\)
Step-by-step Solution
Detailed explanation
\(R = \frac{v^2 \sin(2\theta)}{g}\) \(180\sqrt{3} = \frac{60^2 \sin(2\theta)}{10} \Rightarrow \sin(2\theta) = \frac{180\sqrt{3} \times 10}{3600} = \frac{\sqrt{3}}{2}\) \(2\theta = 60^{\circ}\) or \(2\theta = 180^{\circ} - 60^{\circ} = 120^{\circ}\) \(\theta = 30^{\circ}\) or…
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