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A bomb initially at rest at a height of ...

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

Text Solution

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Initial relative velocity `u_(rel)=10-(-10)=20 m//s`
Relative acceleration t = 2 sec, relative separation.
`S_(rel)=u_(rel)t+(1)/(2)a_(rel)t^(2)=(20xx2)+0=40 m`
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