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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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It is possible to project a particle with a given velocity in two possible ways so as to make it pass through a point at a horizontal distance R from the point of projection . If $t_1$ and $t_2$ are the time taken to reach this point in two possible ways then what is $t_1t_2$?

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Let the angle of projection in the two cases by $\theta_0$ and $(90^{\circ}- \theta_{0})$ i.e $v_0 \sin \theta$ in the second case. Hence,
$R= (v_0 \cos \theta_0)t_1$
$R= (v_0 \sin \theta_0) t_2$
therefore $t_1t_2 =\large\frac{R^2}{v_0 ^2 \sin \theta _0 \cos \theta_0 }$----(i)
But $R= \large\frac{2v_0^2 \sin \theta_0 \cos \theta_0}{g}$-----(ii)
Hence from (i) and (ii)
$t_1t_2=\large\frac{2R}{g}$
answered Jul 15, 2014 by meena.p
edited Aug 13, 2014 by balaji.thirumalai
 

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