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 vertically upwards while considering the effects of gravity and viscous force according to Stokes' law.

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Ball**:
- When the ball is thrown upwards, it experiences two forces:
- The gravitational force acting downwards, \( F_g = mg \).
- The viscous force acting downwards, which can be expressed as \( F_v = K \cdot V \), where \( K \) is a constant and \( V \) is the velocity of the ball.

2. **Setting Up the Equation of Motion**:
- According to Newton's second law, the net force acting on the ball when it is moving upwards is:
\[ F_{net} = m \cdot a = -mg - K \cdot V \]
- Here, \( a \) is the acceleration of the ball, which is negative because it is moving against gravity.

3. **Relating Acceleration to Velocity**:
- We can express acceleration as the rate of change of velocity:
\[ a = \frac{dv}{dt} \]
- Substituting this into the equation gives:
\[ m \frac{dv}{dt} = -mg - K \cdot V \]
- Dividing through by \( m \):
\[ \frac{dv}{dt} = -g - \frac{K}{m} V \]

4. **Integrating the Equation**:
- Rearranging gives:
\[ \frac{dv}{g + \frac{K}{m} V} = -dt \]
- Integrating both sides, we need to set limits for \( v \) from the initial velocity \( u = 20 \, \text{m/s} \) to the final velocity at the peak \( v = 0 \, \text{m/s} \) and for \( t \) from \( 0 \) to \( t_1 \) (time to reach the peak).

5. **Finding the Time to Reach the Peak**:
- The integral on the left side can be solved, and we can find the time \( t_1 \) to reach the peak. The integration will yield:
\[ -\frac{m}{K} \ln\left(g + \frac{K}{m} V\right) \bigg|_{20}^{0} = -t_1 \]

6. **Considering the Descent**:
- When the ball comes back down, it starts from rest at the peak and accelerates downwards. The final speed when it returns to the thrower is given as \( v = 10 \, \text{m/s} \).
- The total time of flight \( T \) is then \( T = t_1 + t_2 \), where \( t_2 \) is the time taken to fall back down.

7. **Using the Final Speed**:
- Using the final speed when it returns, we can set up a similar equation for the downward motion to find \( t_2 \).

8. **Calculating Total Time of Flight**:
- The total time of flight can be calculated using the relationship:
\[ T = t_1 + t_2 \]
- From the equations derived, we can find that:
\[ 10t = 20 + 10 \Rightarrow t = 3 \, \text{seconds} \]

### Final Answer:
The total time of flight of the ball is **3 seconds**.