AP EAMCET · PHYSICS · Motion In Two Dimensions
If the maximum height and range of a projectile are \(3 \mathrm{~m}\) and \(4 \mathrm{~m}\) respectively, then the velocity of the projectile is (Take, \(g=10 \mathrm{~ms}^{-2}\) )
- A \(20 \sqrt{\frac{6}{5}} \mathrm{~ms}^{-1}\)
- B \(10 \sqrt{\frac{3}{2}} \mathrm{~ms}^{-1}\)
- C \(10 \sqrt{\frac{2}{3}} \mathrm{~ms}^{-1}\)
- D \(20 \sqrt{\frac{5}{6}} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(10 \sqrt{\frac{2}{3}} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Here, maximum height, \(H=3 \mathrm{~m}\) and range, \(R=4 \mathrm{~m}\) As, maximum height, \(H=\frac{u^2 \sin ^2 \theta}{2 g}\) and maximum range, \(R=\frac{u^2 \sin 2 \theta}{g}\) then,\(H=3=\frac{u^2 \sin ^2 \theta}{2 g}\) ...(i) and \(R=4=\frac{u^2 \sin 2 \theta}{g}\) The…
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