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…

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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 $t_1$ and $t_2$ are times taken to reach this point in two possible ways, then the product $t_1 t_2$ is proportional to

Aicharya · Accepted Answer

$\begin{aligned} & T_1=\frac{2 u \sin 	heta}{g} \ & 	ext { and } \quad T_2=\frac{2 u \cos 	heta}{g} \ & T_1 T_2=\frac{2 u^2 \sin 	heta \cos 	heta}{g}=\frac{2 r}{g} \ & {\left[\because R=\frac{v^2 \sin 2 	heta}{g}ight]} \ & \end{aligned}$

It is possible to project a particle with a given velocity in two possible ways so as to make them pass…

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