A ball of mass m is thrown upward with a velocity `upsilon`. If air exerts an average resisting force F, the velocity with which the ball returns to the thrower is

Options

A. `vsqrt((mg)/(mg+F))`
B. `vsqrt((F)/(mg+F))`
C. `vsqrt((mg-F)/(mg+F))`
D. None of these

Answer

To solve the problem of finding the velocity with which a ball returns to the thrower after being thrown upward with an initial velocity \( v \) while experiencing an average resisting force \( F \), we can follow these steps:

### Step 1: Analyze the upward motion of the ball
When the ball is thrown upward with an initial velocity \( v \), it will rise until it reaches its maximum height \( H \). At this maximum height, the velocity of the ball becomes zero.

### Step 2: Apply the work-energy principle for the upward motion
Using the work-energy principle, we can equate the initial kinetic energy to the work done against gravity and the resisting force:
\[ \frac{1}{2} mv^2 = mgh + Fh \]
Here, \( mgh \) is the work done against gravity, and \( Fh \) is the work done against the resisting force \( F \). Rearranging this gives:
\[ \frac{1}{2} mv^2 = mgh + Fh \]

### Step 3: Analyze the downward motion of the ball
When the ball falls back down, it will start from rest at height \( H \) and will reach the thrower with some velocity \( v_1 \). The forces acting on the ball during its downward motion are the gravitational force \( mg \) and the resisting force \( F \).

### Step 4: Apply the work-energy principle for the downward motion
For the downward motion, we can write:
\[ \frac{1}{2} mv_1^2 = mgh - Fh \]
Here, \( mgh \) is the work done by gravity, and \( -Fh \) is the work done against the resisting force \( F \).

### Step 5: Set up the equations
From the upward motion, we have:
\[ \frac{1}{2} mv^2 = mgh + Fh \quad \text{(1)} \]
From the downward motion, we have:
\[ \frac{1}{2} mv_1^2 = mgh - Fh \quad \text{(2)} \]

### Step 6: Divide the equations
Dividing equation (1) by equation (2):
\[ \frac{\frac{1}{2} mv^2}{\frac{1}{2} mv_1^2} = \frac{mgh + Fh}{mgh - Fh} \]
This simplifies to:
\[ \frac{v^2}{v_1^2} = \frac{mgh + Fh}{mgh - Fh} \]

### Step 7: Solve for \( v_1^2 \)
Rearranging gives:
\[ v_1^2 = v^2 \cdot \frac{mgh - Fh}{mgh + Fh} \]

### Step 8: Solve for \( v_1 \)
Taking the square root of both sides:
\[ v_1 = v \sqrt{\frac{mgh - Fh}{mgh + Fh}} \]

### Final Result
Thus, the velocity with which the ball returns to the thrower is:
\[ v_1 = v \sqrt{\frac{F - mg}{F + mg}} \]