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, \[{{t}_{1}}{{t}_{2}}\propto R\]You need to login to perform this action.
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