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(C) Let the mass of the body be $2m$. It explodes into two fragments of mass $m$ each. By conservation of linear momentum, the initial momentum is zer

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$2$

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(C) Let the mass of the body be $2m$. It explodes into two fragments of mass $m$ each. By conservation of linear momentum, the initial momentum is zero, so the final momenta must be equal and opposite. If one fragment has velocity $\vec{v}_1 = -10 \hat{i} - gt \hat{j}$, the other must have $\vec{v}_2 = 10 \hat{i} - gt \hat{j}$.The position vectors of the fragments relative to the point of explosion at time $t$ are $\vec{r}_1 = -10t \hat{i} - \frac{1}{2}gt^2 \hat{j}$ and $\vec{r}_2 = 10t \hat{i} - \frac{1}{2}gt^2 \hat{j}$.The angle between the position vectors is $90^{\circ}$, so their dot product is zero:$\vec{r}_1 \cdot \vec{r}_2 = 0$$(-10t \hat{i} - \frac{1}{2}gt^2 \hat{j}) \cdot (10t \hat{i} - \frac{1}{2}gt^2 \hat{j}) = 0$$-100t^2 + \frac{1}{4}g^2t^4 = 0$Since $t 
eq 0$, we divide by $t^2$:$-100 + \frac{1}{4}g^2t^2 = 0$$\frac{1}{4}g^2t^2 = 100$$g^2t^2 = 400$Taking $g = 10 \,ms^{-2}$:$100t^2 = 400$$t^2 = 4$$t = 2 \,s$.

At high altitude, a body at rest explodes into two equal fragments. One fragment receives a horizontal velocity of $10 \,ms^{-1}$. The time taken by the two position vectors connecting the point of explosion to the fragments to make an angle of $90^{\circ}$ is (in $\,s$)

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