A) \[\frac{1}{{{R}^{2}}}\]
B) \[\frac{1}{R}\]
C) R
D) \[{{R}^{2}}\]
Correct Answer: C
Solution :
| [c] We know that range of projectile is same for complementary angles i.e., for\[\theta \]and\[(90{}^\circ -\theta ).\] |
| \[\therefore \] \[{{T}_{1}}=\frac{2u\sin \theta }{g}\] |
| \[{{T}_{2}}=\frac{2u\sin ({{90}^{o}}-\theta )}{g}=\frac{2u\cos \theta }{g}\] |
| and \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] |
| Therefore, \[{{T}_{1}}{{T}_{2}}=\frac{2u\sin \theta }{g}\times \frac{2u\cos \theta }{g}\] |
| \[=\frac{2{{u}^{2}}(2\sin \theta \cos \theta )}{{{g}^{2}}}\] |
| \[=\frac{2{{u}^{2}}(\sin 2\theta )}{{{g}^{2}}}=\frac{2R}{g}\] |
| \[\therefore \] \[{{T}_{1}}{{T}_{2}}\propto R\] |
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