A person of mass `70kg` jumps from a stationary helicopter with the parachute open. As he falls through `50m` height, he gains a speed of `20ms^-1`. The work done by the viscous air drag is

Options

A. (a) `21000J`
B. (b) `-21000J`
C. (c) `-14000J`
D. (d) `14000J`

Explanation

To solve the problem of calculating the work done by the viscous air drag on a person jumping from a helicopter, we can follow these steps:

### Step 1: Identify the known values
- Mass of the person, \( m = 70 \, \text{kg} \)
- Height fallen, \( h = 50 \, \text{m} \)
- Final velocity, \( v = 20 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (assuming standard value)

### Step 2: Calculate the change in kinetic energy
The change in kinetic energy (\( \Delta KE \)) can be calculated using the formula:
\[ \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]
where \( u \) is the initial velocity. Since the person starts from rest, \( u = 0 \):
\[ \Delta KE = \frac{1}{2} \times 70 \times (20)^2 - \frac{1}{2} \times 70 \times (0)^2 \]
\[ \Delta KE = \frac{1}{2} \times 70 \times 400 = 14000 \, \text{J} \]

### Step 3: Calculate the work done by gravity
The work done by gravity (\( W_G \)) is given by:
\[ W_G = mgh \]
Substituting the known values:
\[ W_G = 70 \times 10 \times 50 = 35000 \, \text{J} \]

### Step 4: Apply the work-energy theorem
According to the work-energy theorem:
\[ W_G + W_V = \Delta KE \]
where \( W_V \) is the work done by the viscous air drag. Rearranging gives:
\[ W_V = \Delta KE - W_G \]
Substituting the values we calculated:
\[ W_V = 14000 - 35000 = -21000 \, \text{J} \]

### Step 5: Conclusion
The work done by the viscous air drag is:
\[ W_V = -21000 \, \text{J} \]

### Final Answer
The work done by the viscous air drag is \( -21000 \, \text{J} \).

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