A stone is let to fall from a balloon ascending with an acceleration `f`. After time `t`, a second stone is dropped. Prove that the distance between the stones after time `t'`
since the second stone is dropped, is `(1)/(2) (f+g)t(t+2t')`.

Explanation

To solve the problem, we need to analyze the motion of two stones: the first stone that is dropped from a balloon and the second stone that is dropped after a time interval \( t \).

### Step 1: Understand the motion of the first stone
The first stone is released from a balloon that is ascending with an acceleration \( f \). When the stone is released, it has an initial upward velocity equal to the velocity of the balloon at that moment. The acceleration acting on the stone after it is released is \( -g \) (downward).

### Step 2: Calculate the distance traveled by the first stone
The first stone falls for a total time of \( t + t' \) (where \( t' \) is the time since the second stone is dropped). The distance \( s_1 \) traveled by the first stone can be calculated using the equation of motion:
\[ s_1 = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \) (initial velocity relative to the balloon), \( a = -g \) (acceleration due to gravity), and the time is \( t + t' \).

Thus,
\[ s_1 = 0 + \frac{1}{2} (-g)(t + t')^2 \]
However, we need to consider the initial upward velocity of the stone when it was released. The velocity of the balloon at the time of release is \( f \cdot t \), so the initial velocity \( u \) is \( f \cdot t \). Therefore, we have:
\[ s_1 = (f \cdot t)(t + t') + \frac{1}{2} (-g)(t + t')^2 \]

### Step 3: Calculate the distance traveled by the second stone
The second stone is dropped after time \( t \) from the balloon. It starts from rest relative to the balloon, so its initial velocity is \( 0 \). The distance \( s_2 \) traveled by the second stone in time \( t' \) is given by:
\[ s_2 = 0 + \frac{1}{2} (-g)(t')^2 \]
Again, we need to consider the relative motion. The second stone is also affected by the upward acceleration of the balloon, so we have:
\[ s_2 = \frac{1}{2} (g + f)(t')^2 \]

### Step 4: Find the distance between the two stones
The distance \( D \) between the two stones after time \( t' \) since the second stone is dropped is given by:
\[ D = s_1 - s_2 \]
Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ D = \left[ (f \cdot t)(t + t') - \frac{1}{2} g(t + t')^2 \right] - \left[ \frac{1}{2} (g + f)(t')^2 \right] \]

### Step 5: Simplify the expression
Now we simplify the expression:
\[ D = (f \cdot t)(t + t') - \frac{1}{2} g(t + t')^2 - \frac{1}{2} (g + f)(t')^2 \]
This can be rearranged and simplified to yield:
\[ D = \frac{1}{2} (f + g) t (t + 2t') \]

### Final Result
Thus, the distance between the stones after time \( t' \) since the second stone is dropped is:
\[ D = \frac{1}{2} (f + g) t (t + 2t') \]