Kinematics Question 634
For a stone thrown from a lower of unknown height, the maximum range for a projection speed of 10 m/s is 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} $
BETA
