question_answer

The maximum range of a gun on horizontal terrain is \[16\,\,km\] if\[g=10\,\,m/{{s}^{2}}\]. What must be the muzzle velocity of the shell?              [BHU M-2003]

A)  \[200\,\,m/s\]                 

B)  \[100\,\,m/s\]

C)  \[400\,\,m/s\]                 

D)  \[300\,\,m/s\]

Correct Answer: C

Solution :

                 Key Idea: For maximum horizontal range body should be projected at an angle of\[{{45}^{\circ }}\]. The horizontal range covered is given by \[R={{u}_{x}}\times T\] \[=u\,\,\cos \,\,\theta \times \frac{2u\,\,\sin \,\theta }{g}\]                                        \[={{u}^{2}}\,\,\frac{\left( 2\,\sin \,\theta \,\,\cos \,\theta  \right)}{g}\]                 \[R=\frac{{{u}^{2}}\,\sin \,20}{g}\] For\[{{R}_{\max }}\,\sin \,2\theta =1\],                 \[\therefore \]\[\theta ={{45}^{\circ }}\]                 \[\therefore \]                  \[{{R}_{\max }}=\frac{{{u}^{2}}}{g}\]                 \[\Rightarrow \]                               \[u=\sqrt{{{R}_{\max }}g}\] Given,\[{{R}_{\max }}=16\,km=16\times {{10}^{3}}\,m,\,g=10\,m/{{s}^{2}}\]                                 \[u=\sqrt{16\times {{10}^{3}}\times 10}\]                                 \[u=400\,m/s\] Note: Horizontal range is same whether the body is projected at \[\theta \] or \[\left( {{90}^{\circ }}-\theta  \right)\,i.e.,\]for complementary angles of projection range is same.