• # question_answer If${{t}_{1}}$and${{t}_{2}}$are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then$t_{1}^{2}+t_{2}^{2}$is equal to 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,     $t_{1}^{2}+t_{2}^{2}=\frac{{{(2u\sin \alpha )}^{2}}}{{{g}^{2}}}+\frac{{{(2u\cos \alpha )}^{2}}}{{{g}^{2}}}$ $=\frac{4{{u}^{2}}}{{{g}^{2}}}({{\sin }^{2}}\alpha +{{\cos }^{2}}ga)=\frac{4{{u}^{2}}}{{{g}^{2}}}$

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