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Q. 4.30 A fighter plane flying horizontally at an altitude of 1.5\; km with speed  720\; km/h  passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 \; m \; \; s ^{-1}  to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take\; g=10\; m\; s ^{-2}).

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According to the question the situation is shown below:-  

                                           Motion in a plane ,    20241

Now,  The horizontal distance travelled by the shell  =  Distance travelled by plane

or                                         u \sin \Theta \ t\ =\ vt

or                                              \sin \Theta \ =\ \frac{v}{u}

or                                                             =\ \frac{200}{600}

So,                                                   \Theta \ =\ 19.5^{\circ}

So, the required height will be:- 

                                                   H\ =\ \frac{u^2\ \sin^2(90\ -\ \Theta )}{2g}

or                                                       =\ \frac{600^2\ \cos^2 \Theta }{2g}

or                                                       =\ 16006.48\ m

or                                                       =\ 16\ Km

Posted by

Devendra Khairwa

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