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 upward and then coming back down. We will use the equations of motion under constant acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity of the ball, \( u = 20 \, \text{m/s} \) (upward)
- Final velocity of the ball when it returns to the thrower, \( v = -10 \, \text{m/s} \) (downward)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Use the Equation of Motion:**
We can use the equation of motion that relates initial velocity, final velocity, acceleration, and time:
\[ v = u + at \]
Here, \( a = -g \) (since gravity acts downward).

3. **Set Up the Equation:**
Substituting the known values into the equation:
\[ -10 = 20 - 10t \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ -10t = -10 - 20 \]
\[ -10t = -30 \]
\[ t = \frac{-30}{-10} = 3 \, \text{s} \]

5. **Conclusion:**
The time of flight of the ball is \( t = 3 \, \text{s} \).