A) \[{{u}^{2}}/g\]
B) \[4{{u}^{2}}/{{g}^{2}}\]
C) \[{{u}^{2}}/2g\]
D) 1
Correct Answer: B
Solution :
If two particles having same initial velocity\[u\]and range R, then their direction must be opposite i.e., the direction of projection of them are \[\alpha \] and\[90{}^\circ -\alpha \]. \[\therefore \] \[{{t}_{1}}=\frac{2u\sin \alpha }{g}\] and \[{{t}_{2}}=\frac{2u\sin ({{90}^{o}}-\alpha )}{g},\] \[\Rightarrow \] \[{{t}_{2}}=\frac{2u\cos \alpha }{g}\] Now, \[t_{1}^{2}+t_{2}^{2}=\frac{{{(2u\sin \alpha )}^{2}}}{{{g}^{2}}}+\frac{{{(2u\cos \alpha )}^{2}}}{{{g}^{2}}}\] \[=\frac{4{{u}^{2}}}{{{g}^{2}}}({{\sin }^{2}}\alpha +{{\cos }^{2}}ga)=\frac{4{{u}^{2}}}{{{g}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
