A parachutist after bailing out falls 52 m without friction. When the parachute opens, she decelerates at `2.1 ms^(-2)` & reaches the ground with a speed of `2.9 ms^(-1)`.
(i) How long has been the parachutist in the air? (ii) At what height did the fall begin?

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (i): How long has the parachutist been in the air?

1. **Identify the first part of the fall**:
- The parachutist falls freely under gravity for a distance of \( s_1 = 52 \, \text{m} \).
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).
- We need to find the time \( t_1 \) taken to fall this distance.

2. **Use the equation of motion**:
\[ s = \frac{1}{2} g t_1^2 \]
Substituting the known values:
\[ 52 = \frac{1}{2} \times 10 \times t_1^2 \]
\[ 52 = 5 t_1^2 \]
\[ t_1^2 = \frac{52}{5} = 10.4 \]
\[ t_1 = \sqrt{10.4} \approx 3.22 \, \text{s} \]

3. **Identify the second part of the fall**:
- After the parachute opens, the parachutist decelerates at \( a = -2.1 \, \text{m/s}^2 \).
- The final velocity when reaching the ground \( v = 2.9 \, \text{m/s} \).
- We need to find the initial velocity \( u \) just after the parachute opens.

4. **Calculate the initial velocity \( u \)**:
- Using the equation \( v^2 = u^2 + 2as \), where \( s \) is the distance fallen after the parachute opens.
- We will first need to find \( s_2 \) (the distance fallen after the parachute opens).

5. **Calculate \( s_2 \)**:
- Rearranging the equation:
\[ 2.9^2 = u^2 + 2(-2.1)s_2 \]
- We need to find \( u \) first. We can find \( u \) using the equation of motion for the first part of the fall:
\[ u = g t_1 = 10 \times 3.22 \approx 32.2 \, \text{m/s} \]

6. **Substituting \( u \) back to find \( s_2 \)**:
\[ 2.9^2 = 32.2^2 + 2(-2.1)s_2 \]
\[ 8.41 = 1036.84 - 4.2s_2 \]
\[ 4.2s_2 = 1036.84 - 8.41 \]
\[ 4.2s_2 = 1028.43 \]
\[ s_2 = \frac{1028.43}{4.2} \approx 244.4 \, \text{m} \]

7. **Calculate \( t_2 \)**:
- Now we can find \( t_2 \) using the equation \( v = u + at \):
\[ 2.9 = 32.2 - 2.1t_2 \]
\[ 2.1t_2 = 32.2 - 2.9 \]
\[ 2.1t_2 = 29.3 \]
\[ t_2 = \frac{29.3}{2.1} \approx 13.95 \, \text{s} \]

8. **Total time in the air**:
\[ T = t_1 + t_2 \approx 3.22 + 13.95 \approx 17.17 \, \text{s} \]

### Part (ii): At what height did the fall begin?

1. **Calculate the total height**:
\[ \text{Total height} = s_1 + s_2 = 52 + 244.4 \approx 296.4 \, \text{m} \]

### Final Answers:
(i) The total time the parachutist has been in the air is approximately \( 17.17 \, \text{s} \).

(ii) The height from which the parachutist fell is approximately \( 296.4 \, \text{m} \).