question_answer

    A fixed mortar fires a bomb at an angle of\[53{}^\circ \]above the horizontal with a muzzle velocity of\[80\text{ }m{{s}^{-1}}\]. A tank is advancing directly towards the mortar on level ground at a constant speed of\[\text{5 }m{{s}^{-1}}\]. The initial separation (at the instant mortar is fired) between the mortar and tank so that the tank would be hit is [Take\[g=10\text{ }m{{s}^{-2}}\]]

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 \[S={{S}_{1}}+{{S}_{2}}=678.4\,m\]