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.

Answer

To solve the problem, we need to analyze the motion of the ball thrown upwards and the forces acting on it. The ball is thrown with an initial velocity of \( u = 20 \, \text{m/s} \) and returns with a final velocity of \( v = 10 \, \text{m/s} \). The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \). We will use the equations of motion to find the time of flight.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Ball**:
- When the ball is thrown upwards, it experiences two forces: the gravitational force acting downwards (\( mg \)) and the viscous force (\( F_v \)) acting downwards as well, which is proportional to its velocity due to Stokes' law.
- The net acceleration \( a \) when the ball is moving upwards is given by:
\[ a = g + a_0 \]
where \( a_0 \) is the acceleration due to the viscous force.

2. **Calculate the Time to Reach the Maximum Height**:
- At the maximum height, the final velocity \( v = 0 \, \text{m/s} \).
- Using the first equation of motion:
\[ v = u - (g + a_0)t \]
Substituting the known values:
\[ 0 = 20 - (10 + a_0)t \quad \text{(1)} \]

3. **Calculate the Time to Fall Back Down**:
- When the ball falls back down, it starts from rest at the maximum height and reaches a speed of \( 10 \, \text{m/s} \) when it returns to the thrower's hand.
- The equation of motion for the downward journey is:
\[ v = u + (g - a_0)t \]
Here, \( u = 0 \) (initial velocity at the top), and substituting the known values:
\[ 10 = 0 + (10 - a_0)t \quad \text{(2)} \]

4. **Solve the Two Equations**:
- From equation (1):
\[ (10 + a_0)t = 20 \quad \Rightarrow \quad t = \frac{20}{10 + a_0} \quad \text{(3)} \]
- From equation (2):
\[ (10 - a_0)t = 10 \quad \Rightarrow \quad t = \frac{10}{10 - a_0} \quad \text{(4)} \]

5. **Equate Equations (3) and (4)**:
\[ \frac{20}{10 + a_0} = \frac{10}{10 - a_0} \]
Cross-multiplying gives:
\[ 20(10 - a_0) = 10(10 + a_0) \]
Expanding both sides:
\[ 200 - 20a_0 = 100 + 10a_0 \]
Rearranging:
\[ 200 - 100 = 20a_0 + 10a_0 \quad \Rightarrow \quad 100 = 30a_0 \quad \Rightarrow \quad a_0 = \frac{100}{30} = \frac{10}{3} \, \text{m/s}^2 \]

6. **Substitute \( a_0 \) Back to Find Time \( t \)**:
- Substitute \( a_0 \) back into equation (3):
\[ t = \frac{20}{10 + \frac{10}{3}} = \frac{20}{\frac{30 + 10}{3}} = \frac{20 \times 3}{40} = \frac{60}{40} = 1.5 \, \text{s} \]

7. **Calculate Total Time of Flight**:
- The total time of flight \( T \) is:
\[ T = 2t = 2 \times 1.5 = 3 \, \text{s} \]

### Final Answer:
The time of flight of the ball is \( \boxed{3} \) seconds.