question_answer

A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall, the angle of projection of ball is

A) \[{{\tan }^{-1}}\left( \frac{3}{2} \right)\]

B) \[{{\tan }^{-1}}\left( \frac{2}{3} \right)\]

C)  \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]

D)  \[{{\tan }^{-1}}\left( \frac{3}{4} \right)\]

Correct Answer: B

Solution :

We know that range of a projectile                 \[R=\frac{{{U}^{2}}\sin \,2\theta}{g}\] Here      \[R=6+18=24\]                 \[\therefore \] \[\frac{{{u}^{2}}\sin 2\theta}{g}=24\] ... (i) The equation of trajectory of a projectile                 \[y=x\tan \theta -\frac{g{{x}^{2}}}{2{{U}^{2}}{{\cos }^{2}}\theta }\]                 \[3=6\tan \theta -\frac{36\,g}{2{{U}^{2}}{{\cos }^{2}}\theta }\] ... (ii)                 Fro. ec, (i), \[\frac{g}{{{U}^{2}}}=\frac{\sin 2\theta }{24}\]                 \[=\frac{\sin \theta cos\theta }{12}\]                 Substituting in Eq. (ii), we get                 \[3=6\tan \theta -\frac{3}{2}\]                 \[\tan \theta =\frac{9}{2}\tan \theta \] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}\left( \frac{2}{3} \right)\]