A parachutist drops freely from helicopter for 20 s before the parachute opends out. Then he descends with retardation of `40 m/s^2`. The height of the helicopter from the ground level is 2500m. When he drops out, his velocity on reaching the ground will be

Options

A. `10 ms^(-1)`
B. `10 sqrt 10 ms^(-1)`
C. `20 ms^(-1)`
D. zero

Answer

To solve the problem step by step, we will break it down into parts:

### Step 1: Calculate the distance fallen in the first 20 seconds
The parachutist drops freely from the helicopter for 20 seconds. The distance fallen under gravity can be calculated using the formula:
\[ S = ut + \frac{1}{2}gt^2 \]
where:
- \( u = 0 \) (initial velocity)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 20 \, \text{s} \)

Substituting the values:
\[ S = 0 \cdot 20 + \frac{1}{2} \cdot 10 \cdot (20)^2 \]
\[ S = 0 + 5 \cdot 400 \]
\[ S = 2000 \, \text{m} \]

### Step 2: Calculate the remaining distance to the ground
The total height of the helicopter from the ground is 2500 m. The distance fallen in the first 20 seconds is 2000 m. Therefore, the remaining distance to the ground is:
\[ \text{Remaining distance} = 2500 - 2000 = 500 \, \text{m} \]

### Step 3: Calculate the velocity at the moment the parachute opens
The velocity of the parachutist just before the parachute opens can be calculated using:
\[ v = u + gt \]
where:
- \( u = 0 \) (initial velocity)
- \( g = 10 \, \text{m/s}^2 \)
- \( t = 20 \, \text{s} \)

Substituting the values:
\[ v = 0 + 10 \cdot 20 \]
\[ v = 200 \, \text{m/s} \]

### Step 4: Calculate the final velocity when reaching the ground
After the parachute opens, the parachutist descends with a retardation of \( 40 \, \text{m/s}^2 \). The net acceleration acting on the parachutist will be:
\[ a = g - \text{retardation} = 10 - 40 = -30 \, \text{m/s}^2 \]

Using the third equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 200 \, \text{m/s} \) (initial velocity when parachute opens)
- \( a = -30 \, \text{m/s}^2 \) (acceleration)
- \( s = 500 \, \text{m} \) (remaining distance)

Substituting the values:
\[ v^2 = (200)^2 + 2 \cdot (-30) \cdot 500 \]
\[ v^2 = 40000 - 30000 \]
\[ v^2 = 10000 \]
\[ v = \sqrt{10000} \]
\[ v = 100 \, \text{m/s} \]

### Final Answer
The velocity of the parachutist when reaching the ground will be \( 100 \, \text{m/s} \).

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