To solve the problem, we need to calculate the distance between two stones after a specific time. Here’s a step-by-step breakdown of the solution:
### Step 1: Understand the motion of the first stone
The first stone is dropped from a balloon that is ascending with an acceleration of \(4 \, \text{m/s}^2\). The effective acceleration acting on the stone when it is released is the gravitational acceleration minus the balloon's acceleration.
- **Effective acceleration**:
\[
a_{\text{eff}} = g - a = 10 \, \text{m/s}^2 - 4 \, \text{m/s}^2 = 6 \, \text{m/s}^2
\]
### Step 2: Calculate the distance traveled by the first stone after \(t = 5 \, \text{s}\)
The first stone is in free fall for \(5 \, \text{s}\) (2 seconds before the second stone is dropped, plus 3 seconds after). We can use the equation of motion:
\[
s_1 = ut + \frac{1}{2} a t^2
\]
Where:
- \(u = 0\) (initial velocity)
- \(a = 6 \, \text{m/s}^2\)
- \(t = 5 \, \text{s}\)
Substituting the values:
\[
s_1 = 0 + \frac{1}{2} \cdot 6 \cdot (5)^2 = 3 \cdot 25 = 75 \, \text{m}
\]
### Step 3: Calculate the distance traveled by the second stone after \(t' = 3 \, \text{s}\)
The second stone is dropped after \(2 \, \text{s}\) and falls for \(3 \, \text{s}\). The distance it travels is given by:
\[
s_2 = ut' + \frac{1}{2} a t'^2
\]
Where:
- \(u = 0\)
- \(a = 6 \, \text{m/s}^2\)
- \(t' = 3 \, \text{s}\)
Substituting the values:
\[
s_2 = 0 + \frac{1}{2} \cdot 6 \cdot (3)^2 = 3 \cdot 9 = 27 \, \text{m}
\]
### Step 4: Calculate the distance \(d\) between the two stones
The distance \(d\) between the two stones after \(3 \, \text{s}\) since the second stone was dropped is:
\[
d = s_1 - s_2 = 75 \, \text{m} - 27 \, \text{m} = 48 \, \text{m}
\]
### Step 5: Find \(\frac{d}{28}\)
Now we need to calculate \(\frac{d}{28}\):
\[
\frac{d}{28} = \frac{48}{28} = \frac{12}{7}
\]
### Final Answer
Thus, the value of \(\frac{d}{28}\) is:
\[
\frac{d}{28} = \frac{12}{7}
\]
