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

A bomb initially at rest at a height of 40 m above the ground suddenly

1.	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
Answer» Solution :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`