# How fast would a golf ball fall if dropped from an airplane? In MPH. How long to reach that speed?

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Drag force F:

F = ρa*V^2/2*C*A

being :

C the drag coeff (0.5 for a sphere)

A the cross-section area PI*r^2 in m^2

ρa the density of air equal to 1.25 kg/m^3

V the speed in m/sec

weight force WF = 4/3*PI*r^3*ρb*g (ρb is the density of ball equal to 1250 kg/m^3)

terminal speed is given when such quantities equate , i.e. :

ρa*V^2/2*C*PI*r^2 = 4/3*PI*r^3*ρb*g

since ρb/ρa = 1250/1.25 = 10^3 , g ≅ 10 m/sec^2 and r = 21.4*10^-3 m, then :

V^2/4 = 4/3*21.4*10^-3*10^3*10

V = √ 21.4*16/3*10 ≅ 34 m/sec = 76 MPH

the time needed for reaching that speed is nearly 3 times the time taken in vacuum , i.e. 3*34/10 ≅ 10 sec

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• The ball will start at zero speed downwards, and then ramp up to 72 mph (terminal velocity). It will take about 6.5 seconds to reach that speed. The sideways speed will start with whatever the airplane has, and in 7 to 10 seconds decrease to nearly zero.

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• Depends on the elevation

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