Physics Help Forum How to find terminal velocity

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 Mar 12th 2018, 01:55 PM #1 Junior Member   Join Date: Mar 2018 Posts: 1 How to find terminal velocity DISCLAIMER: I am new to this forum so I am not entirely sure of how to layout and where to post, my post. I am trying to get some physics in a program I am working on, and I want to know how to find the terminal velocity of a perfect sphere in air on Earth. The ball, perfect sphere, has a radius of 4cm (metric system). My code is this: /* Ball Physics Ball has radius of 4cm */ PVector position; float velocity = 0; // Velocity float gravity = 9.80665; // Standard gravity float airDensity = 0.001225; // In grams/cm^3 float mass = 75; // In grams float radius = 4; // In cm's float dragCoefficient = 0.47; // For perfect sphere in air float volume = (4/3)*PI*(radius*radius*radius); // In cm^3 float density = mass/volume; // In grams/cm^3 float projectedArea = PI*radius*radius; // In cm^2 float terminalVelocity = sqrt((2*mass*gravity)/(density*projectedArea*dragCoefficient)); // Terminal velocity NOTE: You don't really have to understand every bit of the code, but just by reading it you should hopefully understand what it's doing. The terminalVelocity at the end is sqrt((2*75*9.8)/(0.001225*48*0.47)) which is 225.45198 I got 225.45198 as the terminal velocity is this correct? If not, what went wrong? Also what would it be measured in, in this case? m/s etc.:. Thanks so much!
 Mar 12th 2018, 10:46 PM #2 Senior Member   Join Date: Apr 2017 Posts: 434 Here's the equation for drag on the falling sphere ..D = Cd * .5 * density * V^2 * A Cd is the drag coeff ..... V is velocity ........ A is area of sphere = pie r squared (r=4) Cd =0.47 (dimensionless ) density (air) =1.225 kg/m2 A= 0.005026 m2... D= 0.47 * 0.5 * 1.225 * 0.005026 * (V squared)..... = 0.001447 * (V squared) At terminal velocity D =mg = 0.075 * 9.80665 = 0.7355 0.7355 = .001447 * (V squared) V = 22.54m/sec The density I calculate for the 4cm sphere is 0.28 .... not very dense , so 22.54 m/sec (50 MPH) sounds about right
Mar 13th 2018, 04:26 AM   #3
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 Originally Posted by MathGuy DISCLAIMER: I am new to this forum so I am not entirely sure of how to layout and where to post, my post. I am trying to get some physics in a program I am working on, and I want to know how to find the terminal velocity of a perfect sphere in air on Earth. The ball, perfect sphere, has a radius of 4cm (metric system). My code is this: /* Ball Physics Ball has radius of 4cm */ PVector position; float velocity = 0; // Velocity float gravity = 9.80665; // Standard gravity float airDensity = 0.001225; // In grams/cm^3 float mass = 75; // In grams float radius = 4; // In cm's float dragCoefficient = 0.47; // For perfect sphere in air float volume = (4/3)*PI*(radius*radius*radius); // In cm^3 float density = mass/volume; // In grams/cm^3 float projectedArea = PI*radius*radius; // In cm^2 float terminalVelocity = sqrt((2*mass*gravity)/(density*projectedArea*dragCoefficient)); // Terminal velocity NOTE: You don't really have to understand every bit of the code, but just by reading it you should hopefully understand what it's doing. The terminalVelocity at the end is sqrt((2*75*9.8)/(0.001225*48*0.47)) which is 225.45198 I got 225.45198 as the terminal velocity is this correct? If not, what went wrong? Also what would it be measured in, in this case? m/s etc.:. Thanks so much!
You need to convert to SI units first. This makes sense, since you're using g = 9.81 m/s$\displaystyle ^2$, not g = 0.0981 cm/s$\displaystyle ^2$.

I get v = 1.4921 m/s by plugging values into that equation with SI units.

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