Bill asked in Science & MathematicsPhysics · 2 months ago

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

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  • oubaas
    Lv 7
    2 months ago
    Favorite Answer

    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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  • 2 months ago

    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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  • 2 months ago

    Depends on the elevation

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