Two stones dropped at the same time. Quick question, .

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When two identical stones are dropped from a tall building one after the other, the vertical distance between them will increase over time. Although both stones fall at the same acceleration of 9.81 m/s², the first stone, released earlier, accelerates faster than the second stone. This results in the first stone traveling at a greater speed, causing the distance between them to grow. The mathematical explanation shows that the difference in height increases as time progresses. Understanding this concept clarifies why the answer is that the distance increases, despite both stones being affected by the same gravitational force.
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Homework Statement


Two identical stones are dropped from a tall building, one after the other. Assume air resistance is negligible. While both stones are falling, what will happen to the vertical distance between them?


Homework Equations



a) It will increase.

b) It will first increase and then remain constant.

c) It will remain the same.

d) It will decrease.




The Attempt at a Solution



Assuming air resistance in negligible, wouldn't the stones both be affected equally and fall to the ground at the same rate of gravitational pull? They would both fall at the rate of 9.81m/s^2 and thus the distance between them would remain the same? Or would it increase until both were falling at this rate and then remain equal?

The book says that the answer is "It will increase" but that would mean that two things being affected by the same force have different accelerations? Why is this?
 
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The key phrase is 'one after the other'. They are not dropped at the same time.
 
wolfmanzak said:
Assuming air resistance in negligible, wouldn't the stones both be affected equally and fall to the ground at the same rate of gravitational pull?
The would experience the same acceleration, yes.
wolfmanzak said:
They would both fall at the rate of 9.81m/s^2
Correct.
wolfmanzak said:
and thus the distance between them would remain the same?
No.
wolfmanzak said:
Or would it increase until both were falling at this rate and then remain equal?
No.

Suppose the first stone was released at t=0 from a height of h=h_0. Its height is then governed by

h_1 = h_0 -\frac{g}{2}t^2

Yes? Suppose the second stone is released from the same height at t=t_0>0. The its height is

h_2 = h_0 - \frac{g}{2}(t-t_0)^2\;\;\;\text{for}\;\;\;t>t_0.

The difference in height is then

h_2-h_1 = -\frac{g}{2}(t-t_0)^2 + \frac{g}{2}t^2

which obviously increases with time (you can plot it to convince yourself).

More intuitively, since the first stone starts accelerating before the second stone and both accelerate at the same rate, the first stone is always traveling faster than the second.

Does that make sense?

Edit: I see Doc beat me to it :frown:
 
I think I understand the concept now. It makes a little more sense after these explanations. Thank you both.
 
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