A) \[\frac{1}{{{R}^{2}}}\]
B) \[\frac{1}{R}\]
C) R
D) \[{{R}^{2}}\]
Correct Answer: C
Solution :
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, \[{{T}_{1}}{{T}_{2}}=\frac{2u\sin \theta }{g}\times \frac{2u\cos \theta }{g}\] \[=\frac{2{{u}^{2}}(2\sin \theta \cos \theta )}{{{g}^{2}}}\] \[=\frac{2{{u}^{2}}(\sin 2\theta )}{{{g}^{2}}}=\frac{2R}{g}\] \[\therefore \] \[{{T}_{1}}{{T}_{2}}\propto R\]
You need to login to perform this action.
You will be redirected in
3 sec
