It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point P at a horizontal distance $r$ from the point of projection. If $t_{1}$ and $t_{2}$ are times taken to reach this point in two possible ways, then the product $t_{1} t_{2}$ is proportional to:

\frac{1}{r}

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r

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r^{2}

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\frac{1}{r^{2}}

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Solution

The correct option is A $r$
The particle will follow two equations:
$r = v c o s θ t$
and $t = \frac{2 v s i n θ}{g}$
$⟹ \frac{r^{2}}{v^{2} t^{2}} + \frac{t^{2} g^{2}}{4 v^{2}} = 1$
$⟹ g^{2} t^{4} - 4 v^{2} t^{2} + 4 r^{2} = 0$

This is a quadratic equation in $t^{2}$ with its solutions $t_{1}^{2}, t_{2}^{2}$ . For a quadratic equation, $a x^{2} + b x + c = 0$ , product of roots is given by $c / a$

$⟹ t_{1}^{2} t_{2}^{2} = \frac{4 r^{2}}{g^{2}}$
$⟹ t_{1} t_{2} = \frac{2 r}{g} \propto r$

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It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point P at a horizontal distance r from the point of projection. If t1 and t2 are times taken to reach this point in two possible ways, then the product t1t2 is proportional to:

The correct option is A rThe particle will follow two equations:r=vcosθtand t=2vsinθg⟹r2v2t2+t2g24v2=1⟹g2t4−4v2t2+4r2=0This is a quadratic equation in t2 with its solutions t21,t22. For a quadratic equation, ax2+bx+c=0, product of roots is given by c/a⟹t21t22=4r2g2⟹t1t2=2rg∝r