# A balloon is ascending vertically with an acceleration of .2 m/s^2 . Two story es are dropped from it at an interval of 2s. What is the distance between them when the second stone is dropped.?

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by mathukkutty

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by mathukkutty

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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

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

==========================

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