A) \[\frac{1}{2}g\,{{t}_{1}}{{t}_{2}}\]
B) \[g\,{{t}_{1}}{{t}_{2}}\]
C) \[\frac{1}{4}g\,{{t}_{1}}{{t}_{2}}\]
D) \[2g\,{{t}_{1}}{{t}_{2}}\]
Correct Answer: A
Solution :
We know that for a given range R there are two Directions of projection, \[viz\,\alpha \]and \[\frac{\pi }{2}-\alpha .\]Thus If \[u\] is the velocity of projection, then \[R=\frac{{{u}^{2}}\sin 2\alpha }{g},{{t}_{1}}=\frac{2u}{g}\sin \alpha \] and \[{{t}_{2}}=\frac{2u\sin \left( \frac{\pi }{2}-\alpha \right)}{g}=\frac{2u\cos \alpha }{g}\] \[\therefore \] \[{{t}_{1}}{{t}_{2}}=\frac{4{{u}^{2}}\sin \alpha \cos \alpha }{{{g}^{2}}}\] \[\Rightarrow \] \[{{t}_{1}}{{t}_{2}}=\frac{2{{u}^{2}}\sin 2\alpha }{g}\] \[\Rightarrow \] \[\frac{1}{2}g\,{{t}_{1}}{{t}_{2}}=\frac{{{u}^{2}}\sin 2\alpha }{g}\] \[\Rightarrow \] \[\frac{1}{2}g\,{{t}_{1}}{{t}_{2}}=R\]
You need to login to perform this action.
You will be redirected in
3 sec
