ઊંચાઇ પર રહેલા સ્થિર પદાર્થના બે સમાન ટુકડ

As the body at rest explodes into two equal parts, they acquire equal velocities in opposite directions according to conservation of momentum.
When the angle between the radius vectors connecting the point of explosion to the fragments is $90^o$, each radius vector makes an angle $45^o$ with the vertical.
To satisfy this condition, the distance of free fall $AD$ should be equal to the horizontal range in same interval of time.
$AD = DB$
$AD = 0 + \frac{1}{2} \times 10{t^2} = 5{t^2}$
$DB = ut = 10t$
 $5{t^2} = 10t \Rightarrow t = 2\sec $