A) 678.4m
B) \[614.4\,\mu \]
C) 64 m
D) None of these
Correct Answer: A
Solution :
The situation is shown clearly in figure. Time of flight of bomb is \[T=\frac{2u\sin \theta }{g}\] \[=\frac{2\times 80\times 4}{10\times 5}=12.85\] Distance travelled by tank in T second is \[{{S}_{1}}=5T=5\times 12.8=64\,m\] The horizontal distance travelled by bomb in T second is \[{{S}_{2}}=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[{{S}_{2}}=\frac{{{80}^{2}}\times 2\times \frac{3}{5}\times \frac{4}{5}}{10}=614.4\,m\] So, required separation \[S={{S}_{1}}+{{S}_{2}}=678.4\,m\]You need to login to perform this action.
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