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(B) Let the particle reach point $P$ at time $t$. The horizontal distance is given by:$x = u \cos 	heta \cdot t$$10 = u \cos 45^{\circ} \cdot t$$10 =

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Let the particle reach point $P$ at time $t$. The horizontal distance is given by:$x = u \cos 	heta \cdot t$$10 = u \cos 45^{\circ} \cdot t$$10 = u \cdot \frac{1}{\sqrt{2}} \cdot t \implies t = \frac{10 \sqrt{2}}{u} \quad \dots (i)$The vertical distance is given by:$y = u \sin 	heta \cdot t - \frac{1}{2} g t^2$$7.5 = u \sin 45^{\circ} \cdot t - \frac{1}{2} \cdot 10 \cdot t^2$$7.5 = u \cdot \frac{1}{\sqrt{2}} \cdot t - 5 t^2 \quad \dots (ii)$Substituting the value of $t$ from Eq. $(i)$ into Eq. $(ii)$:$7.5 = u \cdot \frac{1}{\sqrt{2}} \cdot \left( \frac{10 \sqrt{2}}{u} ight) - 5 \left( \frac{10 \sqrt{2}}{u} ight)^2$$7.5 = 10 - 5 \cdot \frac{100 \cdot 2}{u^2}$$7.5 = 10 - \frac{1000}{u^2}$$\frac{1000}{u^2} = 10 - 7.5 = 2.5$$u^2 = \frac{1000}{2.5} = 400$$u = 20 \ m/s$

Column-$I$ (Angle of projection)	Column-$II$
$A. \theta = 45^{\circ}$	$1. \frac{K_h}{K_i} = \frac{1}{4}$
$B. \theta = 60^{\circ}$	$2. \frac{gT^2}{R} = 8$
$C. \theta = 30^{\circ}$	$3. \frac{R}{H} = 4\sqrt{3}$
$D. \theta = \tan^{-1}(4)$	$4. \frac{R}{H} = 4$

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