AP EAMCET · PHYSICS · Motion In Two Dimensions
A body is projected from the earth at angle \(30^{\circ}\) with the horizontal with some initial velocity. If its range is \(20 \mathrm{~m}\), the maximum height reached by it is : (in metres)
- A \(5 \sqrt{3}\)
- B \(\frac{5}{\sqrt{3}}\)
- C \(\frac{10}{\sqrt{3}}\)
- D \(10 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\(R=\frac{u^2 \sin 2 \theta}{g}\) \(\therefore \quad 20=\frac{u^2 \sin \left(2 \times 30^{\circ}\right)}{g}\) \(\Rightarrow \quad \frac{u^2}{g}=\frac{20}{\sin 60^{\circ}}=\frac{20}{\sqrt{3}} \times 2=\frac{40}{\sqrt{3}}\) Now, \(H=\frac{u^2 \sin ^2 \theta}{2 g}\)…
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