question_answer

A bullet fired at an angle of \[{{30}^{o}}\] with the horizontal hits the ground 3 km away. By adjusting its angle of projection, can one hope to hit a target 5 km away. Assume the muzzle speed to be same and the air resistance is negligible

A)  possible to hit a target 5 km away

B)  not possible to hit a target 5 km away

C)  prediction is not possible

D)  none of the above

Correct Answer: B

Solution :

Key Idea: It is not possible to hit beyond maximum range. The body covers a horizontal distance AB during its flight. This horizontal range is given by \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\]                                      ?(i) where, u is velocity of projection, \[\theta \] is angle of projection and g is acceleration due to gravity. For maximum horizontal range \[\sin 2\theta =1\]                 \[\therefore \]  \[R{{}_{\max }}=\frac{{{u}^{2}}}{g}\]                       ?(ii)                 Given, \[R=3km,\theta ={{30}^{o}}\]                 \[\therefore \]From Eq. (i)                 \[\frac{{{u}^{2}}}{g}=\frac{R}{\sin 2\theta }=\frac{3}{\sin {{60}^{o}}}=\frac{3\times 2}{\sqrt{3}}=\sqrt{3}\times 2\]                                 \[\frac{{{u}^{2}}}{g}=3.464\,m\] Hence, maximum range with velocity of projection u cannot be more than 3.464m. Hence, it is not possible to hit a target 5 km away.