A ball is thrown vertically upwards with a velocity of `49 m//s`. Calulate
(i) The maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.

Explanation

To solve the problem step by step, we will break it down into two parts: (i) calculating the maximum height the ball reaches and (ii) calculating the total time it takes to return to the surface of the Earth.

### Step 1: Calculate the Maximum Height

1. **Identify the known values**:
- Initial velocity (u) = 49 m/s (upwards)
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s² (downwards)

2. **Use the equation of motion**:
We will use the equation:
\[ v^2 = u^2 - 2gh \]
Here, we need to rearrange the equation to solve for height (h):
\[ h = \frac{u^2 - v^2}{2g} \]

3. **Substitute the known values**:
\[ h = \frac{(49)^2 - (0)^2}{2 \times 9.8} \]
\[ h = \frac{2401}{19.6} \]
\[ h \approx 122.5 \text{ meters} \]

### Step 2: Calculate the Total Time Taken to Return to the Surface

1. **Calculate the time to reach maximum height**:
We will use the equation:
\[ v = u - gt \]
Rearranging to solve for time (t):
\[ t = \frac{u - v}{g} \]

2. **Substitute the known values**:
\[ t = \frac{49 - 0}{9.8} \]
\[ t = \frac{49}{9.8} \]
\[ t \approx 5 \text{ seconds} \]

3. **Calculate total time for the entire journey**:
Since the time taken to go up is equal to the time taken to come down, the total time (T) is:
\[ T = t_{\text{up}} + t_{\text{down}} = 5 + 5 = 10 \text{ seconds} \]

### Final Answers:
(i) The maximum height to which the ball rises is approximately **122.5 meters**.
(ii) The total time taken to return to the surface of the Earth is **10 seconds**.