Q1 A BALLOON IS ASCENDING VERTIcally WITH AN ACCELERATION OF .2m\\sec2.two stones are dropped from it at an interval of 2seconds.find the distance between them 1.5 seconds after second stone is released.Q2 the velocity displacement graph of a particle moving along a straight line having negative slope which intersects the y axis(vo ,vo>0) & which intersects the displacements (xo,xo>0) the the most suitable acceleration displacement graph will be what ?

Hi

Ans1: Fix the datum at The initial position off the balloon and assume its velocity to be zero at that time. Therefore the first stone dropped wud move for 3.5 seconds (2 second interval + 1.5 second fall for second stone) downward with acceleration g. Therefore its displacement from the datum is

s1 = 1/2 x g x (3.5)² .

Now, considering the second stone:

It has been dropped wen balloon is at a ht of 2a from its initial position. a= acceleration of balloon. ht = (1/2 xa x 2²) = 2a .

velocity of balloon at this ht = v = ut + 1/2 a t² . t=2 sec, u=0 . therefore v= 2a in upward direction. therefore initial velocity of second stone dropped is 2a in upward direction.  hence calculate the distance travelled by second stone in 1.5 sec by

s2= 2a(1.5) + 1/2x gx (1.5)² .

NOW ANALYSE, whether it is above the datum or below it, then add or substract to get the distance between the two stones.