A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments . One of them starts moving vertically downwards with an initial speed of 10 m /s .If acceleration due to gravity is `10m//s^(2)` , What is the separation between the fragments 2s after the explosion?

Explanation

Case (i) : (for downward moving frament )
Initial velocity , u = 10 `ms^(-1)`
Acceleration , a = +g = `10ms^(-2)`
Time , t = 2s
From the equation of motion , `s=ut+(1)/(2)at^(2)`
the distance moved in downward direction is ,
`s_(1)=10xx2+(1)/(2)xx10xx(2)^(2)=40m`
Case (ii) (for upward moving fragment )
Given that two fragments are identical hence, after explosion the fragments move in opposite directions . Here the first fragment moves in downward direction , hence second fragment moves upward direction .
Again from s = ut + `(1)/(2)"at"^(2)` we can write ,
`s_(2)=-10xx2+(1)/(2)xx10xx4=-20+20=0m`
`therefore` Separation between the fragments 2s
after the explosion = `s_(1)-s_(2)=40-0=40m`