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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 them starts moving vertically downwards with an initial speed 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 `a_("rel") =g -g=0`
After t=2 sec, relative separation
After t=2 sec, relative separation
`S_("rel") =u_(rel) t + 1/2 a_(rel)t^(2) =(20 xx 2) + 0= 40 m`
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