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: Determine the initial conditions
- The parachutist drops from a height of 2500 m.
- The initial velocity (u) when he drops is 0 m/s.
- The time of free fall (t) before the parachute opens is 20 seconds.
- The acceleration due to gravity (g) is approximately 10 m/s².

### Step 2: Calculate the velocity after 20 seconds of free fall
We can use the first equation of motion:
\[ v = u + at \]
Where:
- \( v \) = final velocity after 20 seconds
- \( u \) = initial velocity = 0 m/s
- \( a \) = acceleration = 10 m/s²
- \( t \) = time = 20 s

Substituting the values:
\[ v_1 = 0 + (10 \, \text{m/s}^2 \times 20 \, \text{s}) \]
\[ v_1 = 200 \, \text{m/s} \]

### Step 3: Calculate the distance fallen in 20 seconds
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) = distance fallen
- \( u \) = initial velocity = 0 m/s
- \( a \) = acceleration = 10 m/s²
- \( t \) = time = 20 s

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

### Step 4: Determine the remaining distance to the ground
The total height from the ground is 2500 m, and the parachutist has fallen 2000 m. Therefore, the remaining distance to the ground is:
\[ \text{Remaining distance} = 2500 \, \text{m} - 2000 \, \text{m} = 500 \, \text{m} \]

### Step 5: Analyze the descent after the parachute opens
After the parachute opens, the parachutist descends with a retardation (negative acceleration) of 40 m/s². We can use the third equation of motion to find the final velocity just before hitting the ground.

Let \( v \) be the final velocity just before hitting the ground, and \( v_1 \) is the initial velocity at the moment the parachute opens (200 m/s). The distance to fall after the parachute opens is 500 m.

Using the third equation of motion:
\[ v^2 = v_1^2 + 2as \]
Where:
- \( a = -40 \, \text{m/s}^2 \) (retardation)
- \( s = 500 \, \text{m} \)

Substituting the values:
\[ v^2 = (200)^2 + 2 \times (-40) \times 500 \]
\[ v^2 = 40000 - 40000 \]
\[ v^2 = 0 \]

### Step 6: Calculate the final velocity
Taking the square root:
\[ v = \sqrt{0} = 0 \, \text{m/s} \]

### Conclusion
The velocity of the parachutist when he reaches the ground is 0 m/s.

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