A Ballon starts ascending vertically with an acceleration of 0.2m/s^-2.Two stones are dropped from it at an interval of 2sec.The distance between them 1.5 sec after second stone is released is(g=9.8)

Ans-25m

Suppose when first stone is dropped the upward velocity of the balloon is u. Let at this time height of the balloon be h. We take origin on the ground. The equation of motion of the first stone is y1=h+ut-(1/2)gt^2. After 2s the velocity of the balloon is u+0.2X2=u+0.4. This is initial velocity of second stone in upward direction . The height from ground is h+uX2+(1/2)X0.2X4=h+2u+0.4. This will be initial position of 2nd stone. Then, equation of motion for second stone is y2=h+2u+0.4+(u+0.4)(t-2)-(1/2)g(t-2)^2. Therefore y2-y1 can be found. We have taken time from the dropping of first stone. Therefore we ,while calculating y2-y1, have to take t=3.5s.The answer is 50m with g=10m/s^2. and 49m for g=9.8m/s^2.

Remember: acceleration of 0.2 m/s^2 does not act on stones from the moment of dropping of the stones.

i think your answer must be wrong.