A ballon starts rising from ground from rest at some constant acceleration. After some time, a storne is dropped from it. If the stone reaches the ground in the same time in which balloon reached the dropping poing from ground, find the acceleration of the balloon.

Explanation

To solve the problem, we need to analyze the motion of the balloon and the stone separately. Let's break it down step by step.

### Step 1: Define Variables
Let:
- \( a \) = acceleration of the balloon
- \( t_0 \) = time taken for the balloon to reach the height where the stone is dropped
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))

### Step 2: Calculate the Velocity of the Balloon
The balloon starts from rest and accelerates upwards. The velocity of the balloon after time \( t_0 \) can be calculated using the formula:
\[ v = u + at \]
where \( u = 0 \) (initial velocity). Thus:
\[ v = 0 + a t_0 = a t_0 \]

### Step 3: Calculate the Height of the Balloon
The height \( h \) reached by the balloon after time \( t_0 \) can be calculated using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \):
\[ h = 0 + \frac{1}{2} a t_0^2 = \frac{1}{2} a t_0^2 \]

### Step 4: Analyze the Motion of the Stone
When the stone is dropped from the balloon, it will take the same time \( t_0 \) to reach the ground. The stone has an initial velocity equal to the velocity of the balloon at the moment of release, which is \( a t_0 \).

Using the second equation of motion for the stone:
\[ s = ut + \frac{1}{2} g t^2 \]
Here, \( s \) is the height \( h \) (which is negative since it is falling), \( u = a t_0 \), and \( t = t_0 \):
\[ -h = a t_0 t_0 + \frac{1}{2} g t_0^2 \]
Substituting \( h = \frac{1}{2} a t_0^2 \):
\[ -\frac{1}{2} a t_0^2 = a t_0^2 + \frac{1}{2} g t_0^2 \]

### Step 5: Rearranging the Equation
Rearranging the equation gives:
\[ -\frac{1}{2} a t_0^2 = a t_0^2 + \frac{1}{2} g t_0^2 \]
\[ -\frac{1}{2} a t_0^2 - a t_0^2 = \frac{1}{2} g t_0^2 \]
\[ -\frac{3}{2} a t_0^2 = \frac{1}{2} g t_0^2 \]

### Step 6: Solve for \( a \)
Dividing both sides by \( \frac{1}{2} t_0^2 \):
\[ -3a = g \]
Thus:
\[ a = -\frac{g}{3} \]
Since we are interested in the magnitude of acceleration:
\[ a = \frac{g}{3} \]

### Final Answer
The acceleration of the balloon is:
\[ \boxed{\frac{g}{3}} \]