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

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$r$

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(B) The equation of the trajectory of 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 equation 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 roots be $	an 	heta_1$ and $	an 	heta_2$. Then $	an 	heta_1 	an 	heta_2 = \frac{y + g r^2 / 2 u^2}{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 relation for projectile motion,it is a standard result that $t_1 t_2 = \frac{2 r}{g 	an \alpha}$ where $\alpha$ is the angle of elevation of the point. For a fixed point $(r, y)$,$t_1 t_2 = \frac{2 r}{g 	an 	heta_{elevation}}$.Since $g$ is constant,$t_1 t_2 \propto r$.

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

Similar Questions

Two projectiles of same mass and with same velocity are thrown at an angle $60^o$ and $30^o$ with the horizontal,then which quantity will remain same?

The time of flight of an object projected with speed $20 \, m/s$ at an angle $30^{\circ}$ with the horizontal is .... $s$.

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Match the columns:

Column-$I$ $(R/H_{max})$ Column-$II$ (Angle of projection $\theta$)

$A. 1$ $1. 60^o$

$B. 4$ $2. 30^o$

$C. 4\sqrt{3}$ $3. 45^o$

$D. 4/\sqrt{3}$ $4. \tan^{-1}(4) = 76^o$

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