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(C) Since the body is at rest and explodes into two equal fragments,by the law of conservation of linear momentum,the fragments must move in opposite

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$2$

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(C) Since the body is at rest and explodes into two equal fragments,by the law of conservation of linear momentum,the fragments must move in opposite directions with equal speeds. Let the speed of each fragment be $v = 10 \, m/s$.Let the point of explosion be $A$. After time $t$,the fragments are at positions where their radius vectors from $A$ make an angle of $90^o$. Due to symmetry,each radius vector makes an angle of $45^o$ with the vertical line passing through $A$.Let $D$ be the point directly below $A$ at the same vertical level as the fragments at time $t$. The vertical distance fallen by each fragment is $h = \frac{1}{2}gt^2 = 5t^2$ (taking $g = 10 \, m/s^2$).The horizontal distance traveled by one fragment is $x = vt = 10t$.In the right-angled triangle formed by the vertical displacement and horizontal displacement,the angle with the vertical is $45^o$. Thus,$	an(45^o) = \frac{	ext{horizontal distance}}{	ext{vertical distance}} = \frac{x}{h}$.Since $	an(45^o) = 1$,we have $x = h$.$10t = 5t^2$$5t^2 - 10t = 0 \Rightarrow 5t(t - 2) = 0$.Since $t 
eq 0$,we get $t = 2 \, s$.

At high altitude,a body at rest explodes into two equal fragments,with one fragment receiving a horizontal velocity of $10 \, m/s$. The time taken by the two radius vectors connecting the point of explosion to the fragments to make an angle of $90^o$ is ............ $s$.

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