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A 3 kg box which is 3.9 cm by 3.9 cm by 19.4 cm long is dropped from a plane. What is the ratio of it's terminal velocity while falling with the narrow end down to the terminal velocity if it is dropped with the long end down. Assume that it has a coefficient of drag of 1.12 in either orientation.

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Answer #1

Terminal velocity of a body is given by, V =(2mg/ρAc)^-2

where,

m is mass of the falling object =3 kg

g is the acceleration due to gravity. On Earth this is approximately 9.81 meters per second.

ρ is the density of the fluid the object is falling through= 3kg/ (3.9*3.9*19.4)*10^-6 m^3 = 10166.94 kg/m^3

A is the projected area of the object. This means the area of the object if you projected it onto a plane that was perpendicular to the direction the object is moving = (3.9*3.9)*10^-4 m^2 = 0.001521 m^2

C is the drag coefficient, here 1.12

therefore Terminal velocity, V= {(2*3*9.81)/(10166.94*0.001521*1.12)}^-2 = 1.84349 m/s

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