Coaching Pitchers
By Michael G. Marshall, Ph.D.

Chapter Twenty-Two:    Baseball’s Maximum Falling Velocity

Another way to investigate atmosphere’s effect on airborne objects is their maximum falling velocities.   On average, gravity’s force accelerates falling objects at 32 ft/sec2.   Folklore contends that if someone drops a penny off New York City’s Empire State Building, when it reaches the sidewalk, it can slice through a pedestrian.   This is a gruesome untrue story.   The air molecules of New York City's atmosphere will decelerate the maximum falling velocity of a penny far below the velocity required to penetrate pedestrians.

Suppose someone drops a baseball from 10,000 feet high.   What is baseball’s final velocity when it contacts the player's glove?   The Integrated Final Velocity Formula provides the answer before we factor in atmosphere's effect.

a.    Integrated Final Velocity Formula

vf   =   V(vi2)   +   2(a)(sf -  si)      where V indicates the square root symbol

Where:
1.    vf   stands for final velocity
2.    vi   stands for initial velocity
3.    a   stands for acceleration
4.    sf   stands for final displacement
5.    si   stands for initial displacement

b.    What do we know?

1.    vi   =   0 ft/sec
2.     a   =   g   =   32 ft/sec 2
3.    sf   =   10,000 feet
4.    si   =   0 feet

c.    Falling Baseball Final Velocity Calculation

1.    vf   =   V(vi 2 )   +   2(a)(sf -  si)      where V indicates the square root symbol.
2.          =   V(0)   +   2(32)(10,000   -   (0))
3.          =   V(64000)
4.          =   252.98 ft/sec or 172.45 mph

At 172.45 miles per hour, nobody could catch these baseballs without risking serious injury.   However, baseballs falling in earth’s atmosphere never achieve 172.45 mph.   Air molecules decelerate falling baseballs in the same way that they decelerate pitched baseballs.   What is baseball’s maximum falling velocity? The drag formula provides the answer.

d.    Drag Formula

DG   =   (AD)(SF)(CSA)(RV 2)

Where:
1.    DG   stands for drag
2.    AD   stands for air density
3.    SF   stands for surface friction
4.    CSA  stands for cross-sectional area
5.    RV   stands for relative velocity

e.    Maximum Falling Velocity Formula

Substitute falling object weight (Wt) for drag (DG).   Substitute maximum falling velocity (MFV) for relative velocity (RV).   Rearrange the substituted formula’s terms to have maximum falling velocity by itself.

1. Drag formula                                                                                     DG  =  (AD)(SF)(CSA)(RV 2)
2. Substitute                                                                                          Wt  =  (AD)(SF)(CSA)(MFV 2)
3. Divide both sides by (AD)(SF)(CSA)                  Wt  /  (AD)(SF)(CSA)  =  (AD)(SF)(CSA)(MFV 2 )  /  (AD)(SF)(CSA) 4. Because (AD)(SF)(CSA)/(AD)(SF)CSA) = 1     Wt  /  (AD)(SF)(CSA)  =  MFV2
5. Change sides                                                                              MFV 2   =  Wt  /  (AD)(SF)(CSA)
6. Take the square root of both sides                                         V(MFV 2 )  =  VWt  /  V(AD)(SF)(CSA)
7.   Final answer                                                                                MFV  =  VWt  /  V(AD)(SF)(CSA)

f.    What do we know?

Baseballs weigh 5.25 ounces or 0.328 pounds.   Spheres cross-sectional areas equal the sphere’s circumference squares divided by 3.1416.   Baseball’s circumference is 9.25 inches.   Baseball’s entire cross-sectional area is 27.24 square inches ((9.25)2 / 3.1416).   However, only one-half of falling spheres collide with air molecules.   Therefore, divide baseball’s total cross-sectional area by 2 (13.62 square inches).   Air density (AD) equals 0.00001.   Surface friction (SF) equals 0.7.

1.     Wt  =  0.328 lbs.
2.     AD  =  0.00001
3.     SF  =  0.7
4.    CSA  =  13.62 sq. in.

g.    Maximum Falling Velocity Calculation

1.    MFV   =   Wt   /   V(AD)(SF)(CSA)
2.    MFV   =  0.328   /   V(0.00001)(0.7)(13.62)
3.    MFV   =  V(3441.7628)
4.    MFV   =  58.67 ft/sec or 40 mph

Baseball’s maximum falling velocity is 40 miles per hour.   All fielders should easily catch 40 mph baseballs dropped from any height from which they can see them.   If, with the 32 ft/sec 2 force of gravity continuously accelerating falling baseballs through the air molecules of atmosphere, the maximum velocity baseballs achieve is forty miles per hour, then what influence do air molecules have on pitched baseballs with release velocities of 90 miles per hour and no force continuously accelerating them?   In Chapter 2, I demonstrated that my fastball decelerated almost 13 miles per hour from release to the catcher's glove.   Air molecules caused that deceleration.

FREE BOOK!!!                          Chapter Twenty-Three