Question

a ballon is ascend ing vertically with an accleration of 1m/s. two stone are dropped from it. at a interval of 2 second. find the distance with 1.5 second after the second stone is released.

Answer 1

The balloon is ascending upward with velocity 1m/s.

So , The velocities of dropped stones is reduced by 1m/s because it will be opposite to the direction of stone's motion.

As stones are released with initial velocity 0 wrt balloon
, there initial velocity wrt ground will be -1m/s.

The second stone is released after 2second of first stone, so after 1.5 second, the first stone has been released 2+1.5=3.5sec.

By using the formula..

s=ut + 1/2 gt^2
s= -3.5 + 5(3.5)^2
s=-3.5 + 5×12.25
s=-3.5 +61.25
s=57.75m