question_answer

A projectile can have the same range R for two angles of projection. If\[{{t}_{1}}\]and\[{{t}_{2}}\]are the times of flights in the two cases, then the product of the two times of flights is proportional to       AIEEE  Solved  Paper-2005  

A) \[{{R}^{2}}\]                      

B) \[\frac{1}{{{R}^{2}}}\]   

C)        \[\frac{1}{R}\]                  

D)        \[R\]

Correct Answer: D

Solution :

A projectile can have same range if angles of projection are complementary i.e.,\[\theta \]and\[(90{}^\circ -\theta )\]. Thus, in both cases \[{{t}_{1}}=\frac{2u\sin \theta }{g}\]                                       .?...(i) \[{{t}_{2}}=\frac{2u\sin ({{90}^{o}}-\theta )}{g}\] \[=\frac{2u\,\cos \theta }{g}\]                                                   ??..(ii) From Eqs. (i) and (ii), we get \[{{t}_{1}}{{t}_{2}}=\frac{4{{u}^{2}}\sin \theta \cos \theta }{{{g}^{2}}}\] \[{{t}_{1}}{{t}_{2}}=\frac{2{{u}^{2}}\sin 2\theta }{{{g}^{2}}}\]        \[(\because \sin 2\theta =2\sin \theta \cos \theta )\] \[=\frac{2}{g}\frac{{{u}^{2}}\sin 2\theta }{g}\] \[\therefore \,\,{{t}_{1}}{{t}_{2}}=\frac{2R}{g}\]                               \[\left( \because \,\,R=\frac{{{u}^{2}}\sin 2\theta }{g} \right)\] Hence,  \[{{t}_{1}}{{t}_{2}}\propto R\]