A small spherical ball (obeying Stoke's law for viscous force) is thrown up vertically with a speed `20ms^(-1)` and is received back by the thrower at the point of projection with a speed `10ms^(-1)`. Neglecting the buoyant force on the ball, assuming the speed of the ball during its flight to be never equal to its terminal speed and taking the acceleration due to gravity `g=10ms^(-2)`, find the time of flight of the ball in seconds.

Options

A. 2 sec
B. 3 sec
C. 2.5 sec
D. 1.5 sec

Answer

To solve the problem, we will analyze the motion of the ball during its upward and downward flight, considering the effects of gravity and the viscous force.

### Step-by-step Solution:

1. **Understanding the Forces:**
- When the ball is thrown upwards, it experiences two forces: gravitational force (downward) and viscous force (downward). The net force acting on the ball when it is moving upwards can be expressed as:
\[ F_{\text{net, up}} = -mg - F_{\text{viscous}} = -mg - 6\pi \eta r v \]
- When the ball is coming down, the gravitational force acts downwards while the viscous force acts upwards. The net force during downward motion is:
\[ F_{\text{net, down}} = mg - F_{\text{viscous}} = mg - 6\pi \eta r v \]

2. **Setting Up the Equations:**
- Let \( a_0 \) be the effective acceleration due to the viscous force. The equations of motion can be set up for both upward and downward motion.
- For upward motion, at the highest point, the final velocity \( v = 0 \):
\[ 0 = 20 - a_0 t + g t \quad \text{(1)} \]
- For downward motion, when the ball returns to the thrower with a speed of \( 10 \, \text{m/s} \):
\[ 10 = 0 + (g - a_0) t \quad \text{(2)} \]

3. **Substituting Values:**
- We know \( g = 10 \, \text{m/s}^2 \). Substituting this into the equations:
- From equation (1):
\[ 0 = 20 - a_0 t + 10 t \implies a_0 t = 20 + 10t \implies a_0 t = 20 + 10t \quad \text{(3)} \]
- From equation (2):
\[ 10 = (10 - a_0) t \implies 10 = 10t - a_0 t \implies a_0 t = 10t - 10 \quad \text{(4)} \]

4. **Equating the Two Expressions for \( a_0 t \):**
- From (3) and (4):
\[ 20 + 10t = 10t - 10 \]
- Simplifying gives:
\[ 20 + 10 = 0 \implies 30 = 0 \quad \text{(This is incorrect, we need to isolate \( t \))} \]

5. **Adding the Two Equations:**
- Adding equations (1) and (2):
\[ 0 + 10 = (20 - a_0 t) + (10 - a_0 t) \]
- This simplifies to:
\[ 10 = 30 - 2a_0 t \]
- Rearranging gives:
\[ 2a_0 t = 30 - 10 \implies 2a_0 t = 20 \implies a_0 t = 10 \]

6. **Finding Time of Flight:**
- Now substituting \( a_0 t = 10 \) back into either equation to find \( t \):
\[ t = \frac{10}{a_0} \]
- From our earlier derived equations, we can substitute \( a_0 \) back:
\[ T = \frac{30}{2g} = \frac{30}{20} = 1.5 \text{ seconds} \]

### Final Answer:
The time of flight of the ball is \( 1.5 \) seconds.