Estimate the terminal speed of a wooden sphere

AI Thread Summary
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.
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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:
 
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