Terminal Velocity Of A Falling Baseball 39h33

Terminal Velocity Of a Falling Baseball By William Greco Warrington, Pa. July 2009

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Executive Summary When a baseball falls through air, it’s inertia is opposed by air friction. The ball will achieve a terminal velocity when the mass of the ball matches the air friction forces which oppose it. To reach the terminal velocity the ball must have reached a minimum altitude above the local elevation. The following equations, charts and examples explore the “Terminal Velocity Of a Falling Baseball”. Main The mass (W) of the baseball must match the aerodynamic forces D (air friction) to reach terminal velocity. W=D (equation-1) W = mass of the baseball D = aerodynamic drag Drag is represented by: p D = CD   V 2 A (equation-2) 2 where: CD = Coefficient of drag (0.43 for a sphere) p = air density taken at sea level V = Velocity A = surface area of the sphere (1/2 of the total area) Substituting W for D we have: p W = CD   V 2 A 2

(equation-2a)

Terminal Velocity Of a Falling Baseball By William Greco Warrington, Pa. July 2009

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Solving for V (velocity): CD P  V 2 A p 2 W = CD   V A = 2 2 Multiply both sides by 2: 2W =

CD P  V 2 A

(2) =

2W = CD P  V 2 A

2 Divide both sides by CD P  A: C D  P  V2 A 2W = = CD P  A CD P  A

V2 =

2W C D P  A

Take the square root of both sides : V=

2W CD P  A

(equation-3)

Characteristics of a Baseball Table-1 Variable Metric System Inch-Pound System CD 0.43 0.43 W 1.39 newtons 10.05 poundals 3 P 1.226 kg/m 0.0766 lbs/ft3 A 0.00877 m2 0.0944 ft2 Note: A represents ½ of the baseball sphere because only half of the sphere represents the frontal area that is acted upon by the air resistance.

Terminal Velocity Of a Falling Baseball By William Greco Warrington, Pa. July 2009

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A poundal is a unit of force equal to the force that imparts an acceleration of 1 foot/sec/sec to a mass of 1 pound; equal to 0.1382 newtons

1 newton (N) is defined as the force required to give a mass of 1 kilogram an acceleration of 1 m s-2. 1 pound of force = 4.45 newtons, average baseball weighs = 5 oz 5 oz 1.39 = 0.313 lbs 4.45 = 1.39 newtons = 10.05 poundals oz 0.1382 16 lb Applying table-1 and equation-3 the terminal velocity of a baseball in the metric (SI) system International : V=

2W = CD P  A

24.521

(

2 1.39  m ) = 24.521 0.43 1.226  0.00877 sec

m ft = 80.44 = 54.84 mph sec sec

Applying table-1 and equation-3 the terminal velocity of a baseball in the Inch-Pound (IP) system :

V=

2W = CD P  A

80.4

ft = 54.8 mph sec

(

2 10.05  ft ) = 80.4 0.43 0.0766  0.0944 sec

Terminal Velocity Of a Falling Baseball By William Greco Warrington, Pa. July 2009

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The baseball must achieve a minimum altitude or it will not be able to reach terminal velocity on the way down. The following expressions allow the calculation of this minimum altitude.

V 2 = U 2 + 2gS

(equation-4)

where: S = distance g = gravitational constant U = initial velocity gt 2 S= (equation-5) 2 where: S = distance g = gravitational constant t = time in seconds 2

Substituting 80.4 ft_sec for V into equation-4: 80.4  = U 2 + 2gS 2

80.4 

= 6464  6464 = U 2 + 2gS

subtract U 2 from both sides = 6464-U 2 = U 2 + 2gS - U 2  6464-U 2 = 2gS Divide both sides by 2g :

6464-U 2 2gS 6464-U 2  =   S= (equation-6) 2g 2g 2g

U = 0 at the point where the ball begins to drop, and g = 32.2 ft_sec 2 6464-0 2 S= = 100.4 feet 2 32.2 

The baseball must be at least 100.4 feet above the ground to reach it’s terminal velocity.

Terminal Velocity Of a Falling Baseball By William Greco Warrington, Pa. July 2009

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To calculate the time that it will take for a baseball to reach it’s terminal velocity during a fall from the height that was calculated on page 4 of 6, equation-5 must be used. gt 2 S= solving for t (time) : 2 gt 2 multiply both sides by 2  2S = 2  2 2S gt 2 2S and divide both sides by g   =  t2 = this is what you get g g g and

take the square root of both sides  t =

2S g

plugging in our g = 32.2 ft_sec 2 and distance of 100.4 ft altitude

t=

2 100.4  32.2

= 2.497 seconds to fall from 100.4 feet and reach terminal velocity

Terminal Velocity Characteristics of a Baseball Terminal Velocity 80.4 ft-second Minimum Altitude to required to reach Terminal Velocity 100.4 feet Time from altitude to reach terminal velocity 2.497 seconds When the baseball reaches the ground if traveling at it’s terminal velocity it will have a force of :

80.4

ft 5 oz x = 25.1 foot pounds sec 16 oz lb

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William Greco- 2404 Greensward N. Warrington Pa. [email protected]