A ball of mass 100 gm is projected vertically upwards from the ground with a velocity of `49m//sec`. At the same time another identical ball is dropped from a height of 98 m to fall freely along the same path as that followed by the first ball. After some time the two balls collide and stick together and finally fall to the ground. Find the time of flight of the masses.

Explanation

To solve the problem, we need to analyze the motion of both balls step by step, calculate the time of flight until they collide, and then determine the total time of flight after they stick together.

### Step 1: Calculate the time until the two balls collide

1. **Identify the motion of the first ball (upward motion)**:
- Mass of the ball, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Initial velocity, \( u_1 = 49 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \) (acting downward)

The distance traveled by the first ball after time \( t \) can be calculated using the equation of motion:
\[ h_1 = u_1 t - \frac{1}{2} g t^2 \]

2. **Identify the motion of the second ball (downward motion)**:
- The second ball is dropped from a height of \( h = 98 \, \text{m} \) with an initial velocity of \( u_2 = 0 \).

The distance traveled by the second ball after time \( t \) is given by:
\[ h_2 = \frac{1}{2} g t^2 \]

3. **Set up the equation for collision**:
Since the two balls collide when the total distance covered by both balls equals the height from which the second ball was dropped:
\[ h_1 + h_2 = 98 \]
Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ u_1 t - \frac{1}{2} g t^2 + \frac{1}{2} g t^2 = 98 \]
Simplifying this gives:
\[ 49t = 98 \]
Therefore, solving for \( t \):
\[ t = \frac{98}{49} = 2 \, \text{s} \]

### Step 2: Calculate the height at which they collide

4. **Calculate the height \( h_1 \) at the time of collision**:
\[ h_1 = u_1 t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ h_1 = 49 \times 2 - \frac{1}{2} \times 9.8 \times (2^2) \]
\[ h_1 = 98 - 19.6 = 78.4 \, \text{m} \]

### Step 3: Calculate the velocities of both balls at the time of collision

5. **Calculate the velocity of the first ball at collision**:
\[ v_1 = u_1 - g t = 49 - 9.8 \times 2 = 29.4 \, \text{m/s} \]

6. **Calculate the velocity of the second ball at collision**:
\[ v_2 = g t = 9.8 \times 2 = 19.6 \, \text{m/s} \]

### Step 4: Calculate the combined velocity after collision

7. **Calculate the momentum before collision**:
Since both balls have the same mass \( m \):
\[ \text{Momentum before collision} = m v_1 - m v_2 = m(29.4 - 19.6) = m \cdot 9.8 \]

8. **Calculate the velocity after collision**:
The total mass after collision is \( 2m \):
\[ \text{Momentum after collision} = 2m v \]
Setting the momenta equal:
\[ m \cdot 9.8 = 2m v \implies v = \frac{9.8}{2} = 4.9 \, \text{m/s} \]

### Step 5: Calculate the time taken to reach the highest point and fall back down

9. **Calculate the time to reach the highest point**:
\[ v_f = 0 \quad \text{(at the highest point)} \]
Using the equation:
\[ v_f = v - g t' \implies 0 = 4.9 - 9.8 t' \implies t' = \frac{4.9}{9.8} = 0.5 \, \text{s} \]

10. **Calculate the height reached after collision**:
\[ h' = \frac{v^2}{2g} = \frac{(4.9)^2}{2 \times 9.8} = 1.22 \, \text{m} \]

11. **Calculate the total height from which the combined mass falls**:
\[ h'' = h_1 + h' = 78.4 + 1.22 = 79.62 \, \text{m} \]

12. **Calculate the time to fall from height \( h'' \)**:
Using the equation:
\[ h'' = \frac{1}{2} g t''^2 \implies 79.62 = \frac{1}{2} \cdot 9.8 \cdot t''^2 \]
Solving for \( t'' \):
\[ t'' = \sqrt{\frac{79.62 \times 2}{9.8}} \approx 4 \, \text{s} \]

### Step 6: Calculate the total time of flight

13. **Total time of flight**:
\[ \text{Total time} = t + t' + t'' = 2 + 0.5 + 4 = 6.5 \, \text{s} \]

### Final Answer:
The total time of flight of the masses is **6.5 seconds**.