Question

A piece of hail has a mass of 0.08 kilograms. It presents an area of 0.0014 square meters, and has a drag coefficient of 0.4. Air density near the ground is 1.225 kilograms per cubic meter. What is the hail’s terminal velocity as it approaches the ground?

V t=2wρACd−−−−√


(1 point)
Responses

47.7 meters per second
47.7 meters per second

2,274 meters per second
2,274 meters per second

15.3 meters per second
15.3 meters per second

30.2 meters per second

Answer 1

To calculate the terminal velocity \( V_t \) of the hailstone, we can use the formula:

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

where:
- \( w \) is the weight of the hailstone (which is \( mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \)),
- \( \rho \) is the air density (1.225 kg/m³),
- \( A \) is the cross-sectional area (0.0014 m²),
- \( C_d \) is the drag coefficient (0.4).

First, we calculate the weight \( w \):

\[
w = mg = 0.08 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.7848 \, \text{N}
\]

Now we can plug in the values into the terminal velocity formula:

\[
V_t = \sqrt{\frac{2 \times 0.7848 \, \text{N}}{1.225 \, \text{kg/m³} \times 0.0014 \, \text{m²} \times 0.4}}
\]

Calculating the denominator:

\[
1.225 \times 0.0014 \times 0.4 = 0.000685 \, \text{kg/m·s²}
\]

Calculating the entire expression inside the square root:

\[
\frac{2 \times 0.7848}{0.000685} \approx 2292.566
\]

Now taking the square root:

\[
V_t \approx \sqrt{2292.566} \approx 47.9 \, \text{m/s}
\]

Since this calculated value rounds to approximately 47.7 m/s, the correct answer among the choices given is:

**47.7 meters per second.**