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A 560-g squirrel with a cross-section area facing the fluid of 144 , {cm}² falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a 56-kg person hitting the ground, assuming no drag contribution in such a short distance?

A. Terminal velocity of the squirrel is 3.0 , {m/s}, and the velocity of the person is 21 , {m/s}.
B. Terminal velocity of the squirrel is 21 , {m/s}, and the velocity of the person is 3.0 , {m/s}.
C. Terminal velocity of the squirrel is 10 , {m/s}, and the velocity of the person is 15 , {m/s}.
D. Terminal velocity of the squirrel is 15 , {m/s}, and the velocity of the person is 10 , {m/s}.

1 Answer

4 votes

Final answer:

To estimate the terminal velocity of the squirrel, we can use the formula v = sqrt((2mg)/(ρAC)).

With the given values, the terminal velocity of the squirrel is estimated to be approximately 11.3 m/s.

However, without the cross-sectional area and drag coefficient for the person, we cannot accurately estimate their velocity without drag contribution.

Step-by-step explanation:

To estimate the terminal velocity of the squirrel, we need to use the formula:

v = sqrt((2mg)/(ρAC))

Where:
- v is the terminal velocity
- m is the mass of the squirrel (in kg)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- ρ is the density of air (approximately 1.225 kg/m^3)
- A is the cross-sectional area facing the fluid (in m^2)
- C is the drag coefficient (approximately 1 for a horizontal skydiver)

Calculating with the given values:
- m = 0.56 kg
- A = 0.144 cm^2 = 0.144 x 10^-4 m^2
- C = 1

Plugging the values into the formula, we get
v = sqrt((2 * 0.56 * 9.8) / (1.225 * 0.144 x 10^-4))

v ≈ 11.3 m/s

To estimate the velocity of the person, we need to use the formula:

v = sqrt((2mg)/(ρAC))

Calculating with the given values:
- m = 56 kg (Note: Convert grams to kilograms)
- A = ? (not provided in the question)
- C = ? (not provided in the question)

Since the question does not provide the cross-sectional area and the drag coefficient for the person, we cannot accurately estimate the velocity without this information. Therefore, we cannot determine the velocity of the person hitting the ground without drag contribution.

User Kevin Cantu
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