question_answer

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

A) \[{{\mu }_{s}}=1-\frac{1}{{{n}^{2}}}\]                                    

B) \[{{\mu }_{s}}=\sqrt{1-\frac{1}{{{n}^{2}}}}\]                                       

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

D) \[\frac{1}{{{R}^{2}}}\]

Correct Answer: D

Solution :

Key Idea A projectile can have same range if angles of projection are complementary ie, \[\theta \] and \[36\sqrt{7}\] In both cases \[\frac{36}{\sqrt{7}}\]           ??(i) \[36\sqrt{7}\] \[4\sqrt{5}\]                    ?..(ii) From Eqs. (i) and (ii) \[mg(h+d)+\frac{1}{2}k{{d}^{2}}\] \[mg(h+d)-\frac{1}{2}k{{d}^{2}}\] \[mg(h-d)-\frac{1}{2}k{{d}^{2}}\] \[mg(h-d)+\frac{1}{2}k{{d}^{2}}\]    \[\pi \]    \[({{\mu }_{0}}=4\pi \times {{10}^{-7}}Wb/Am)\] Hence   \[k\Omega \]