Question

A skydiver has a mass of 95 kilograms. Before he releases his parachute, he presents an area of 0.54 meters, and has a drag coefficient of 0.7. Air density at the start of his jump is 0.9 kilograms per cubic meter.
What is the skydiver's terminal velocity before he releases his parachute? Vt =
2w
(1 point)
pACa
77.82 meters per second
73.98 meters per second
23.63 meters per second
6,056.2 meters per second

Answer 1

To calculate the terminal velocity of the skydiver before he releases his parachute, we can use the following formula:

\[
V_t = \sqrt{\frac{2mg}{\rho A C_d}}
\]

Where:
- \( V_t \) = terminal velocity
- \( m \) = mass of the skydiver (95 kg)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( \rho \) = air density (0.9 kg/m³)
- \( A \) = cross-sectional area (0.54 m²)
- \( C_d \) = drag coefficient (0.7)

Now we can substitute the values into the formula:

1. Calculate the weight (\( mg \)):
\[
mg = 95 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 931.95 \, \text{N}
\]

2. Now substitute everything into the terminal velocity formula:
\[
V_t = \sqrt{\frac{2 \times 931.95 \, \text{N}}{0.9 \, \text{kg/m}^3 \times 0.54 \, \text{m}^2 \times 0.7}}
\]

3. Calculate the denominator:
\[
0.9 \times 0.54 \times 0.7 = 0.3402
\]

4. Now, substitute into the terminal velocity expression:
\[
V_t = \sqrt{\frac{1863.90}{0.3402}} = \sqrt{5485.827} \approx 73.95 \, \text{m/s}
\]

Thus, the terminal velocity of the skydiver before he releases the parachute is approximately \( 73.95 \, \text{m/s} \).

This rounds to **73.98 meters per second**.

So the answer is:
**73.98 meters per second**.