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(D) For the bodies to collide,their $x$ and $y$ coordinates must be equal at the same time $t$.Let the first body be $A$ and the second be $B$.For bod

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$0.2$

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(D) For the bodies to collide,their $x$ and $y$ coordinates must be equal at the same time $t$.Let the first body be $A$ and the second be $B$.For body $A$: $x_A = (10 \cos 30^\circ)t = 10 \cdot \frac{\sqrt{3}}{2} \cdot t = 5\sqrt{3}t$ and $y_A = (10 \sin 30^\circ)t - \frac{1}{2}gt^2 = 5t - 5t^2$ (taking $g = 10 \ ms^{-2}$).For body $B$: $x_B = (\sqrt{3}-1) + (v \cos 45^\circ)t = (\sqrt{3}-1) + \frac{v}{\sqrt{2}}t$ and $y_B = (v \sin 45^\circ)t - \frac{1}{2}gt^2 = \frac{v}{\sqrt{2}}t - 5t^2$.Equating $y_A = y_B$: $5t - 5t^2 = \frac{v}{\sqrt{2}}t - 5t^2 \implies 5 = \frac{v}{\sqrt{2}} \implies v = 5\sqrt{2} \ ms^{-1}$.Equating $x_A = x_B$: $5\sqrt{3}t = (\sqrt{3}-1) + \frac{5\sqrt{2}}{\sqrt{2}}t \implies 5\sqrt{3}t = \sqrt{3}-1 + 5t$.$t(5\sqrt{3}-5) = \sqrt{3}-1 \implies t(5(\sqrt{3}-1)) = \sqrt{3}-1$.$t = \frac{\sqrt{3}-1}{5(\sqrt{3}-1)} = \frac{1}{5} = 0.2 \ s$.

Two bodies are projected from points $(0,0)$ and $(\sqrt{3}-1,0)$ with velocities $10 \ ms^{-1}$ and $v \ ms^{-1}$ respectively,as shown in the figure. The time after which they collide in space is . . . . . . . (in $s$)

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