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(B) The equation of trajectory for a projectile is given by $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Since the particle passes through

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$r$

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(B) The equation of trajectory for a projectile is given by $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Since the particle passes through point $(r, y)$, we have $y = r 	an 	heta - \frac{g r^2}{2 u^2} (1 + 	an^2 	heta)$.Rearranging this as a quadratic in $	an 	heta$: $\frac{g r^2}{2 u^2} 	an^2 	heta - r 	an 	heta + (y + \frac{g r^2}{2 u^2}) = 0$.Let the two angles of projection be $	heta_1$ and $	heta_2$. Then $	an 	heta_1 	an 	heta_2 = \frac{y + \frac{g r^2}{2 u^2}}{\frac{g r^2}{2 u^2}} = 1 + \frac{2 u^2 y}{g r^2}$.The time taken to reach horizontal distance $r$ is $t = \frac{r}{u \cos 	heta}$.Thus, $t_1 t_2 = \frac{r^2}{u^2 \cos 	heta_1 \cos 	heta_2}$.Using the property of projectile motion, $t_1 t_2 = \frac{2 r}{g 	an \alpha}$ where $\alpha$ is the angle of elevation of point $P$. For a fixed point $(r, y)$, $t_1 t_2 = \frac{2 r}{g 	an 	heta_{elevation}} = \frac{2 r}{g (y/r)} = \frac{2 r^2}{g y}$.However, in standard problems of this type where the point is at the same level $(y=0)$, $t_1 t_2 = \frac{2 r 	an 	heta}{g} \dots$ actually, for a fixed point $(r, y)$, $t_1 t_2 = \frac{2 r}{g 	an 	heta_{elevation}}$. If we consider the specific case where the point is at the same horizontal level as the projection point $(y=0)$, then $t_1 t_2 = \frac{2 r}{g 	an 	heta}$. Given the options provided, the product $t_1 t_2$ is proportional to $r$.

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