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A BALLOON IS MOVING VERTICALLY UPWARD WITH CONSTANT ACCN (g/2) in upward direction. The particle A' was dropped from the balloon and 2sec later another particle B' was dropped from the same balloon. Assuming that the motion of the balloon remains unaffected find the separation distance between A' and B', 6sec after dropping particle B'. none of the particles reaches the ground during the time interval.
I thnk one more parameter is needed that is the unitial velocity of baloon when particle A was dropped.....
let the velocity of baloon when particle A was dropped is Ua
2 seconds later velocity of baloon will be v=u+at
Ub=Ua+g2/2= Ua+g
initial velocity of first particle A will be equal to Ua as the particle is placed on baloon and
initial velocity of second particle B will be equal to Ub
now, displacement of particle A after 8 seconds will be Sa=8Ua-64g/2
Sa=8Ua-32g.......1
displacement of particle B after 6 seconds is Sb=6Ub-36g/2
Sb=6(Ua+g) -18g
Sb=6Ua-12g..............2
Sb-Sa=20g-2Ua..........
if Ua (initial velocity of baloon) is known then Sb-Sa can be calculated......
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