question_answer

A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18m 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 \[y=x\tan \theta -\frac{g{{x}^{2}}}{2{{U}^{2}}{{\cos }^{2}}\theta }\] \[3=6\tan \theta -\frac{36g}{2{{U}^{2}}{{\cos }^{2}}\theta }\]                     ? (ii)                 Form Eq. (i), \[\frac{g}{{{U}^{2}}}=\frac{\sin 2\theta }{24}\] \[=\frac{\sin \theta \cos \theta }{12}\] Substituting in Eq. (ii), we get \[3=6\tan \theta -\frac{3}{2}\] \[\tan \theta =\frac{9}{2}\tan \theta \] \[\Rightarrow \]                               \[\theta ={{\tan }^{-1}}\left( \frac{2}{3} \right)\]