After jumping out from the plane, a parachutist falls 80 m without friction. When he opens up the parachute, he decelerates at 2 m `s^(-2)`. He reaches the ground with a speed of 4 m `s^(-1)`. How long did the parachutist spend his time in the air? (Take g = 10 `m s^(-2)`)

Options

A. 4 s
B. 16s
C. 18 s
D. 22 s

Explanation

To solve the problem, we need to break it down into two parts: the free fall before the parachute opens and the descent after the parachute opens.

### Step 1: Calculate the time taken during free fall (t1)

1. **Given data**:
- Distance fallen before opening parachute (s1) = 80 m
- Initial velocity (u) = 0 m/s (since he jumps from rest)
- Acceleration due to gravity (g) = 10 m/s²

2. **Using the equation of motion**:
\[ s = ut + \frac{1}{2}gt^2 \]
Substituting the known values:
\[ 80 = 0 \cdot t_1 + \frac{1}{2} \cdot 10 \cdot t_1^2 \]
This simplifies to:
\[ 80 = 5t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{80}{5} = 16 \]
Taking the square root:
\[ t_1 = 4 \text{ seconds} \]

### Step 2: Calculate the velocity just before the parachute opens (v_i)

1. **Using the equation of motion**:
\[ v = u + gt \]
Substituting the known values:
\[ v_i = 0 + 10 \cdot 4 = 40 \text{ m/s} \]

### Step 3: Calculate the time taken after the parachute opens (t2)

1. **Given data**:
- Final velocity (v) = 4 m/s (when he reaches the ground)
- Initial velocity (v_i) = 40 m/s (just before the parachute opens)
- Deceleration (a) = -2 m/s² (negative because it's deceleration)

2. **Using the equation of motion**:
\[ v = v_i + at \]
Substituting the known values:
\[ 4 = 40 - 2t_2 \]
Rearranging gives:
\[ 2t_2 = 40 - 4 = 36 \]
Thus:
\[ t_2 = \frac{36}{2} = 18 \text{ seconds} \]

### Step 4: Calculate the total time spent in the air

1. **Total time (T)**:
\[ T = t_1 + t_2 = 4 + 18 = 22 \text{ seconds} \]

### Final Answer:
The total time the parachutist spent in the air is **22 seconds**.

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