A man, 65 kg, descends to the ground with the help of a parachute, 18 kg. The parachute is hemispherical in shape, 2m diameter. Density of air can be taken as 0.00125 g/cm3 and its kinematic viscosity as 0.15 stoke. What is the terminal velocity of the parachute?(Take CD = 1.5 and g = 1000 cm/sec2)a)16.7 m/secb)15.8 m/secc)15.0 m/secd)14.1 m/secCorrect answer is option 'A'. Can you explain this answer?

Question

Answer

Understanding Terminal Velocity
Terminal velocity is the constant speed an object eventually reaches when falling through a fluid (in this case, air). At this speed, the gravitational force acting on the object is balanced by the drag force produced by the fluid.
Given Data
- Mass of the man, $ m = 65 \, \text{kg} $
- Mass of the parachute, $ M = 18 \, \text{kg} $
- Total mass, $ W = m + M = 83 \, \text{kg} $
- Diameter of the parachute, $ D = 2 \, \text{m} $ (Radius $ r = 1 \, \text{m} $)
- Density of air, $ \rho = 0.00125 \, \text{g/cm}^3 = 1.25 \, \text{kg/m}^3 $
- Drag coefficient, $ C_D = 1.5 $
- Gravitational acceleration, $ g = 1000 \, \text{cm/sec}^2 = 10 \, \text{m/sec}^2 $
Drag Force Calculation
The drag force $ F_d $ can be calculated using the formula:
$$
F_d = \frac{1}{2} C_D \rho A v^2
$$
Where:
- $ A $ is the cross-sectional area of the parachute.
The area $ A $ of a hemisphere is given by:
$$
A = \pi r^2 = \pi (1)^2 = \pi \, \text{m}^2
$$
Weight Calculation
The weight $ W $ acting downwards is:
$$
W = (m + M)g = 83 \times 10 = 830 \, \text{N}
$$
Setting Forces Equal at Terminal Velocity
At terminal velocity, the drag force equals the weight:
$$
\frac{1}{2} C_D \rho A v_t^2 = W
$$
Substituting the values:
$$
\frac{1}{2} \times 1.5 \times 1.25 \times \pi \times v_t^2 = 830
$$
Solving for $ v_t $:
1. Calculate the left side:
- $ \frac{1}{2} \times 1.5 \times 1.25 \approx 0.9375 $
- Area $ A \approx 3.14 \, \text{m}^2 $
- Thus, $ 0.9375 \times 3.14 \approx 2.940625 $
2. Set up the equation:
- $ 2.940625 \cdot v_t^2 = 830 $
3. Solve for $ v_t $:
- $ v_t^2 \approx \frac{830}{2.940625} \approx 282.38 $
- $ v_t \approx \sqrt{282.38} \approx 16.8 \, \text{m/sec} $
Thus, the terminal velocity is approximately 16.7 m/sec, confirming option A is

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