A balloon a ascending vertically with an acceleration of `0.4m//s^(2)`. Two stones are dropped from it at an interval of `2 sec.` Find the distance between them `1.5sec.` after the second stone is released. `(g=10m//sec^(2))`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the motion of the stones
The balloon is ascending with an acceleration of \(0.4 \, \text{m/s}^2\). When the first stone is dropped, it has the same upward velocity as the balloon at that moment. The second stone is dropped 2 seconds later.

### Step 2: Calculate the time for each stone
- For the first stone, the total time after it is dropped until we measure the distance is \(2 \, \text{s} + 1.5 \, \text{s} = 3.5 \, \text{s}\).
- For the second stone, the total time after it is dropped is \(1.5 \, \text{s}\).

### Step 3: Calculate the distance traveled by the first stone (S1)
Using the equation of motion:
\[ S_1 = ut + \frac{1}{2} a t^2 \]
Where:
- \(u = 0\) (initial velocity of the stone with respect to the balloon when dropped)
- \(a = g - a_{\text{balloon}} = 10 \, \text{m/s}^2 - 0.4 \, \text{m/s}^2 = 9.6 \, \text{m/s}^2\)
- \(t = 3.5 \, \text{s}\)

Substituting the values:
\[ S_1 = 0 \cdot 3.5 + \frac{1}{2} \cdot 9.6 \cdot (3.5)^2 \]
Calculating \(S_1\):
\[ S_1 = \frac{1}{2} \cdot 9.6 \cdot 12.25 = 58.8 \, \text{m} \]

### Step 4: Calculate the distance traveled by the second stone (S2)
Using the same equation:
\[ S_2 = ut + \frac{1}{2} a t^2 \]
Where:
- \(u = 0\)
- \(a = 9.6 \, \text{m/s}^2\)
- \(t = 1.5 \, \text{s}\)

Substituting the values:
\[ S_2 = 0 \cdot 1.5 + \frac{1}{2} \cdot 9.6 \cdot (1.5)^2 \]
Calculating \(S_2\):
\[ S_2 = \frac{1}{2} \cdot 9.6 \cdot 2.25 = 10.8 \, \text{m} \]

### Step 5: Calculate the distance between the two stones
The distance \(d\) between the two stones after 1.5 seconds from the release of the second stone is given by:
\[ d = S_1 - S_2 \]
Substituting the values:
\[ d = 58.8 \, \text{m} - 10.8 \, \text{m} = 48 \, \text{m} \]

### Final Answer
The distance between the two stones 1.5 seconds after the second stone is released is \(48 \, \text{m}\).

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