A balloon is ascending vertically with an acc of 0.2m/s2.Two stones are dropped from it at an interval of 2 sec.Find the distance between them 1.5 sec after the second stone is released(g=9.8)
21net acceleration=9.8-0.2=9.6
now, distance between stones= 1/2*9.6*(3.5)2- 1/2*9.6*(1.5)2
so,distance=49mts
1after the second stone is released
separation of the stones at this instant of dropping of the second stone =
1/2*10*2*2=20 m
discussing motion from the frame of the second stone .
rel. velocity of 1 stone = v12= 9.8*2+2.2*2 = 20 m/s
sepration of the stones after 1.5 s= 20*1.5+20= 50m
