At high altitude, a body explodes at rest into two equal fragments with one fragment receiving horizontal velocity of $10 \,m/s$. Time taken by the two radius vectors connecting point of explosion to fragments to make $90^o $ is ............ $\mathrm{s}$
At high altitude, a body explodes at rest into two equal fragments with one fragment receiving horizontal velocity of $10 \,m/s$. Time taken by the two radius vectors connecting point of explosion to fragments to make $90^o $ is ............ $\mathrm{s}$
$10$
$4$
$2$
$1$
At high altitude, a body explodes at rest into two equal fragments with one fragment receiving horizontal velocity of $10 \,m/s$. Time taken by the two radius vectors connecting point of explosion to fragments to make $90^o $ is ............ $\mathrm{s}$
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 $
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