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. `21000J`
B. `-21000J`
C. `-14000J`
D. `14000J`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the change in kinetic energy (ΔKE)
The change in kinetic energy is given by the formula:
\[ \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]
where:
- \( m = 70 \, \text{kg} \) (mass of the person)
- \( v = 20 \, \text{m/s} \) (final velocity)
- \( u = 0 \, \text{m/s} \) (initial velocity)

Substituting the values:
\[ \Delta KE = \frac{1}{2} \times 70 \times (20)^2 - \frac{1}{2} \times 70 \times (0)^2 \]
\[ \Delta KE = \frac{1}{2} \times 70 \times 400 \]
\[ \Delta KE = 35 \times 400 = 14000 \, \text{J} \]

### Step 2: Calculate the work done by gravity (Wg)
The work done by gravity is given by the formula:
\[ W_g = mgh \]
where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 50 \, \text{m} \) (height fallen)

Substituting the values:
\[ W_g = 70 \times 10 \times 50 \]
\[ W_g = 7000 \times 50 = 35000 \, \text{J} \]

### Step 3: Apply the work-energy theorem
According to the work-energy theorem:
\[ W_{\text{total}} = \Delta KE \]
This can be expressed as:
\[ W_g + W_d = \Delta KE \]
where \( W_d \) is the work done by the viscous drag force.

Rearranging gives:
\[ W_d = \Delta KE - W_g \]

### Step 4: Substitute the values
Substituting the values we calculated:
\[ W_d = 14000 - 35000 \]
\[ W_d = -21000 \, \text{J} \]

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