Question

A balloon is ascending vertically with an acceleration of 0.2 metre per second . Two stone are dropped at t = 0 and at t = 2 second . Find the distance between them 1.5 sec after the second stone is released.


Please answer if you know Properly​

Answer 1

Explanation:

Let us assume that the balloon C continues to rise at an acceleration of "a" = 2 m/s² after  the 1st stone is dropped at t =0 sec.  The first stone A has the same initial velocity upwards as the 2nd stone B at t = 0 sec.  So relative velocity between the two stones is 0.

Acceleration of B after t = 0, is = a = 2 m/s²   wrt ground

Acceleration of A after t = 0,  is = -g = -10m/s²  wrt ground.

Relative to 2nd stone B acceleration of A = (a+g)

In 2 seconds, the separation between A and B = s 

            s = ut+1/2 at² = 0 * 2 + 1/2 (g+a) 2² = 1/2 (2+10)2² = 24 meters

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Alternately, we can assume an initial velocity u of the balloon at t =0, and the stone is dropped at t = t1 and the second stone is dropped at t1 + 2 sec.

Vertical position of stone 1 at t1+2 sec = ( u t1+1/2 a t1² ) + (u + a t1) * 2 - 1/2 g 2² 

Vertical position of stone 2 at t1+2 sec = u (t1 + 2) + 1/2 a (t1+2)² 

Difference between them = 1/2 (a+g) 2² = 1/2 * 12 * 4 = 24 meters

Answer 2

Explanation:

Your answer refer in Attachment

Thank you