1. Sep 12, 2011

### wolfmanzak

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

a) It will increase.

b) It will first increase and then remain constant.

c) It will remain the same.

d) It will decrease.

3. 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?

2. Sep 12, 2011

### Staff: Mentor

The key phrase is 'one after the other'. They are not dropped at the same time.

3. Sep 12, 2011

### Hootenanny

Staff Emeritus
The would experience the same acceleration, yes.
Correct.
No.
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 travelling faster than the second.

Does that make sense?

Edit: I see Doc beat me to it

4. Sep 12, 2011

### wolfmanzak

I think I understand the concept now. It makes a little more sense after these explanations. Thank you both.