Estimate the terminal speed of a wooden sphere

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The discussion revolves around estimating the terminal speed of a wooden sphere with a density of 0.870 g/cm³, a radius of 8.50 cm, and a drag coefficient of 0.500, while considering the density of air as 1.20 kg/m³. The calculated mass of the sphere is 2.24 kg, and the cross-sectional area is 0.0227 m². The terminal velocity formula applied yields a result of 56.8 m/s, but the user reports discrepancies with WebAssign, indicating they are within 10% of the correct answer. Other participants confirm they are obtaining the same result, suggesting a potential issue with the input or parameters used in the calculation. The conversation highlights the need for verification of calculations or possible adjustments to the parameters for accuracy.
MetallicaFNum
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Homework Statement


(a) Estimate the terminal speed of a wooden sphere (density 0.870 g/cm3) falling through air, if its radius is 8.50 cm and its drag coefficient is 0.500. (The density of air is 1.20 kg/m3.)

(b) From what height would a freely falling object reach this speed in the absence of air resistance?


Homework Equations


m = p_sphere*V
A = pi*r^2
v_t= sqrt(2mg/Dp_airA)


The Attempt at a Solution


m = (870 kg/m^3)(4/3)pi(0.085 m)^3 = 2.24 kg
A = pi(0.085 m)^2 = 0.0227 m^2

v_t = sqrt[2(2.24 kg)(9.80 m/s^2)/(0.500)(1.20 kg/m^3)(0.0227 m^3)] = 56.8 m/s

WebAssign keeps telling me I'm within 10% of the correct answer. What am I doing wrong so that I will get the correct answer?

Thanks,
Mark
 
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I don't know, I get your same result.:confused:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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