Question

Question
A ball falls from a tall bridge. The ball has a mass of 0.15 kg. The air density is 1.225 kg/m3
. The ball’s surface area is 0.0026 m2
. The ball’s coefficient of drag is 0.007. What is the ball’s terminal velocity?(1 point)
Responses

30.4 m/s
30.4 m/s

164 m/s
164 m/s

363 m/s
363 m/s

402 m/s

Answer 1

To find the terminal velocity of a falling ball, we can use the following formula for terminal velocity \(v_t\):

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

where:
- \(m\) = mass of the object (kg)
- \(g\) = acceleration due to gravity (\(9.81 \, m/s^2\))
- \(\rho\) = air density (\(kg/m^3\))
- \(C_d\) = coefficient of drag
- \(A\) = surface area of the object (\(m^2\))

Now, substituting the values provided:

- \(m = 0.15 \, kg\)
- \(g = 9.81 \, m/s^2\)
- \(\rho = 1.225 \, kg/m^3\)
- \(C_d = 0.007\)
- \(A = 0.0026 \, m^2\)

Plugging in the values:

\[
v_t = \sqrt{\frac{2 \times 0.15 \times 9.81}{1.225 \times 0.007 \times 0.0026}}
\]

Calculating the numerator:

\[
2 \times 0.15 \times 9.81 = 2.943
\]

Calculating the denominator:

\[
1.225 \times 0.007 \times 0.0026 \approx 0.000022573
\]

Now calculating \(v_t\):

\[
v_t = \sqrt{\frac{2.943}{0.000022573}} \approx \sqrt{130175.8678} \approx 360.5 \, m/s
\]

Rounding gives us approximately \(363 \, m/s\).

Thus, the terminal velocity is \(\boxed{363 \, m/s}\).