Kinematics Question 634

Question: For a stone thrown from a lower of unknown height, the maximum range for a projection speed of 10 m/s it’s obtained for a projection angle of $ 30{}^\circ . $ The corresponding distance between the foot of the lower and the point of landing of the stone is

Options:

A) $ 10m $

B) $ ~20m $

C) $ ( \text{20/}\sqrt{3} )\text{ m}~ $

D)$ ( 10/\sqrt{3} )\text{ m} $

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Answer:

Correct Answer: D

Solution:

[d] $ \tan \theta =\frac{{{u^{2}}}}{\text{Rg}} $

$ \Rightarrow \text{R}\text{=}\frac{{{u^{2}}}}{g\tan \theta }=\frac{100}{10\times \sqrt{3}}=\frac{10}{\sqrt{3}}\text{ m} $



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