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(C) Let the mass of the body be $2m$. After the explosion,it splits into two fragments of mass $m$ each.Since the initial momentum is zero,the final m

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$240\sqrt{3} 	ext{ m}$

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(C) Let the mass of the body be $2m$. After the explosion,it splits into two fragments of mass $m$ each.Since the initial momentum is zero,the final momentum must also be zero. If one fragment has velocity $\vec{v}_1 = 10\sqrt{3} \hat{i}$,the other must have $\vec{v}_2 = -10\sqrt{3} \hat{i}$.At time $t$,the positions of the fragments are $\vec{r}_1 = (10\sqrt{3}t) \hat{i} - (\frac{1}{2}gt^2) \hat{j}$ and $\vec{r}_2 = (-10\sqrt{3}t) \hat{i} - (\frac{1}{2}gt^2) \hat{j}$.The displacement vectors are $\vec{r}_1$ and $\vec{r}_2$. The angle between them is $60^{\circ}$,so $\cos(60^{\circ}) = \frac{\vec{r}_1 \cdot \vec{r}_2}{|\vec{r}_1| |\vec{r}_2|}$.$\frac{1}{2} = \frac{-(10\sqrt{3}t)^2 + (\frac{1}{2}gt^2)^2}{(10\sqrt{3}t)^2 + (\frac{1}{2}gt^2)^2}$.Let $x = 10\sqrt{3}t$ and $y = \frac{1}{2}gt^2 = 5t^2$. Then $\frac{1}{2} = \frac{y^2 - x^2}{x^2 + y^2} \implies x^2 + y^2 = 2y^2 - 2x^2 \implies 3x^2 = y^2 \implies y = \sqrt{3}x$.Substituting $x = 10\sqrt{3}t$ and $y = 5t^2$: $5t^2 = \sqrt{3}(10\sqrt{3}t) = 30t \implies t = 6 	ext{ s}$.The horizontal distance is $|x_1 - x_2| = |10\sqrt{3}t - (-10\sqrt{3}t)| = 20\sqrt{3}t = 20\sqrt{3}(6) = 120\sqrt{3} 	ext{ m}$.Wait,checking the geometry: the angle between vectors is $60^{\circ}$. The horizontal separation is $20\sqrt{3}t$. At $t=6$,distance is $120\sqrt{3}$. Re-evaluating: The angle between $\vec{r}_1$ and $\vec{r}_2$ is $60^{\circ}$. $	an(30^{\circ}) = \frac{x}{y} = \frac{10\sqrt{3}t}{5t^2} = \frac{2\sqrt{3}}{t}$. Since $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$,$t = 6$. The horizontal distance is $20\sqrt{3}t = 120\sqrt{3}$. Given the options,let's re-verify the angle condition. If the angle between displacement vectors is $60^{\circ}$,the horizontal distance is $20\sqrt{3}t$. Calculation yields $120\sqrt{3}$. None of the options match exactly,but $120\sqrt{3}$ is the derived result.

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