A) 0.3 m
B) 2m
C) 1 m
D) 100
Correct Answer: A
Solution :
At the time of throwing body velocity gained by man is \[{{v}_{0}}=\frac{3\times 8}{70}=\frac{24}{70}m/s\] Stopping distance is \[S=\frac{v_{0}^{2}}{2\,\mu g}\] \[s=\frac{24\times 24}{70\times 70\times 2\times 0.02\times 10}\] \[=\frac{576}{49\times 40}=\frac{144}{490}=0.3\,m\]
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