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, 9 is angle of projection and g is acceleration due to gravity. For maximum horizontal range \[\sin \,2=1\] \[\therefore \] \[{{R}_{\max }}=\frac{{{u}^{2}}}{g}\] ?. (ii) Given, \[R=3\,km,\,\theta ={{30}^{o}}\]                 \[\therefore \] From Eq. (i) \[\frac{{{u}^{2}}}{g}=\frac{R}{\sin \,2\theta }=\frac{3}{\sin \,{{60}^{o}}}=\frac{3\times 2}{\sqrt{3}}=\sqrt{3}\times 2\] \[\therefore \] \[\frac{{{u}^{2}}}{g}=3.464\,\,m\] Hence, maximum range with velocity of projection u cannot be more than 3.464 m. Hence, it is not possible to hit a target 5 km away.