A balloon with mass `m` is descending down with an acceleration a `(where altg)` . How much mass should be removed from it so that it starts moving up with an acceleration a?

Options

A. `(2ma)/(g+a)`
B. `(2ma)/(g-a)`
C. `(ma)/(g+a)`
D. `(ma)/(g-a)`

Explanation

To solve the problem step-by-step, we will analyze the forces acting on the balloon in both scenarios: when it is descending and when it is ascending.

### Step 1: Analyze the initial condition (descending)
The balloon with mass \( m \) is descending with an acceleration \( a \). The forces acting on the balloon are:
- Weight \( W = mg \) acting downward.
- Air resistance \( R \) acting upward.

According to Newton's second law, the net force \( F \) acting on the balloon can be expressed as:
\[ F = W - R = ma \]
Substituting the weight, we have:
\[ mg - R = ma \]

### Step 2: Rearranging the equation
From the equation \( mg - R = ma \), we can express the air resistance \( R \):
\[ R = mg - ma \]

### Step 3: Analyze the final condition (ascending)
Now, we want the balloon to ascend with the same acceleration \( a \). If we remove a mass \( x \) from the balloon, the new mass of the balloon becomes \( m - x \). The forces acting on the balloon in this case are:
- New weight \( W' = (m - x)g \) acting downward.
- Air resistance \( R \) still acting upward.

The net force in this scenario will be:
\[ R - (m - x)g = (m - x)a \]

### Step 4: Rearranging the new equation
From the equation \( R - (m - x)g = (m - x)a \), we can express it as:
\[ R = (m - x)g + (m - x)a \]

### Step 5: Equating the two expressions for air resistance
Since the air resistance \( R \) is the same in both scenarios, we can set the two expressions for \( R \) equal to each other:
\[ mg - ma = (m - x)g + (m - x)a \]

### Step 6: Expanding and simplifying
Expanding the right-hand side:
\[ mg - ma = mg - xg + ma - xa \]
Now, we can cancel \( mg \) from both sides:
\[ -ma = -xg + ma - xa \]
Rearranging gives:
\[ -ma + xg + xa = ma \]
\[ xg + xa = 2ma \]

### Step 7: Factoring out \( x \)
Factoring out \( x \) from the left-hand side:
\[ x(g + a) = 2ma \]

### Step 8: Solving for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{2ma}{g + a} \]

### Conclusion
The mass that should be removed from the balloon so that it starts moving up with an acceleration \( a \) is:
\[ x = \frac{2ma}{g + a} \]