AP EAMCET · PHYSICS · Motion In Two Dimensions
If the velocity at the maximum height of a projectile projected at an angle of \(45^{\circ}\) is \(20 \mathrm{~m} \mathrm{~s}^{-1}\), then the maximum height reached by the projectile is (Acceleration due to gravity \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) )
- A 10 m
- B 20 m
- C 30 m
- D 40 m
Answer & Solution
Correct Answer
(B) 20 m
Step-by-step Solution
Detailed explanation
\(v_x = u \cos \theta\) \(20 = u \cos 45^{\circ}\) \(u = \frac{20}{\cos 45^{\circ}} = \frac{20}{1/\sqrt{2}} = 20\sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) \(H = \frac{u^2 \sin^2 \theta}{2g}\) \(H = \frac{(20\sqrt{2})^2 \sin^2 45^{\circ}}{2 \times 10}\)…
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