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 2005]

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

B) \[\frac{1}{{{R}^{2}}}\]  

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

D) \[R\]

Correct Answer: D

Solution :

[d] 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

\[{{t}_{1}}{{t}_{2}}=\frac{4{{u}^{2}}\sin \theta \cos \theta }{{{g}^{2}}}\]

\[{{t}_{1}}{{t}_{2}}=\frac{2{{u}^{2}}\sin 2\theta }{{{g}^{2}}}\]        \[(\because \sin 2\theta =2\sin \theta \cos \theta )\]

\[=\frac{2}{g}\frac{{{u}^{2}}\sin 2\theta }{g}\]

\[\therefore \]

Hence,