A hot- air balloon consists of a basket, one passenger, and some carge. Let the total mass be M. Even though there is an upward lift force on the balloon, the balloon is initially acceleration downwards at a rate of `g//3`.
(a) Draw a free-body diagram for the descending balloon.
(b) Find the upward lift force in terms of the initial total weight Mg.
(c ) The passenger notices that he is heading straight for a waterfall and decides he needs to go up. What fraction of the total weight must he drop overboard so that the balloon accelerates upward at a rate of `g//2`? Assume that the upward lift force remains the same.

Explanation

Let's solve the problem step by step.

### Step 1: Draw the Free-Body Diagram for the Descending Balloon

To draw the free-body diagram, we need to identify the forces acting on the hot-air balloon.

1. **Weight (W)**: The downward force due to gravity acting on the total mass \( M \) of the balloon, which is \( W = Mg \).
2. **Lift Force (F)**: The upward force acting on the balloon, which we will denote as \( F \).

Since the balloon is descending with an acceleration of \( \frac{g}{3} \), the net force acting on the balloon can be expressed as:

\[ \text{Net Force} = \text{Weight} - \text{Lift Force} = Mg - F \]

According to Newton's second law, this net force is also equal to the mass times the acceleration:

\[ Mg - F = M \left(-\frac{g}{3}\right) \]

### Step 2: Find the Upward Lift Force in Terms of the Initial Total Weight \( Mg \)

From the equation we derived in Step 1, we can rearrange it to find the lift force \( F \):

\[ Mg - F = -\frac{Mg}{3} \]

Rearranging gives:

\[ F = Mg + \frac{Mg}{3} = \frac{3Mg}{3} + \frac{Mg}{3} = \frac{4Mg}{3} \]

Thus, the upward lift force \( F \) in terms of the initial total weight \( Mg \) is:

\[ F = \frac{4}{3} Mg \]

### Step 3: Determine the Fraction of Total Weight to Drop for Upward Acceleration of \( \frac{g}{2} \)

Now, we need to find out how much weight the passenger must drop to achieve an upward acceleration of \( \frac{g}{2} \).

Let \( m_1 \) be the new mass of the balloon after dropping some weight. The forces acting on the balloon now are:

1. **Weight**: \( m_1 g \) (downward)
2. **Lift Force**: \( F \) (upward, which we found to be \( \frac{4}{3} Mg \))

Using Newton's second law again, we have:

\[ F - m_1 g = m_1 \left(\frac{g}{2}\right) \]

Substituting \( F \):

\[ \frac{4}{3} Mg - m_1 g = m_1 \left(\frac{g}{2}\right) \]

Rearranging gives:

\[ \frac{4}{3} Mg = m_1 g + m_1 \left(\frac{g}{2}\right) \]

Factoring out \( m_1 g \):

\[ \frac{4}{3} Mg = m_1 g \left(1 + \frac{1}{2}\right) = m_1 g \left(\frac{3}{2}\right) \]

Now, solving for \( m_1 \):

\[ m_1 = \frac{4}{3} \cdot \frac{2}{3} M = \frac{8M}{9} \]

The mass that needs to be dropped \( m_d \) is:

\[ m_d = M - m_1 = M - \frac{8M}{9} = \frac{1M}{9} \]

### Final Answer

The fraction of the total weight \( Mg \) that must be dropped is:

\[ \text{Fraction} = \frac{m_d}{M} = \frac{\frac{1M}{9}}{M} = \frac{1}{9} \]