Copyright 2003
Coaching Pitchers
By Michael G. Marshall, Ph.D.

     Chapter Twenty:   Gravity

     Gravity accelerates all airborne objects downward at 32 ft/sec2.   When hunters fire their rifles perfectly horizontal, the bullets fall to the ground in the same time interval that they fall when dropped from the same height.   Horizontal velocities have no effect on gravity’s downward acceleration force.   How much does gravity downwardly accelerate pitches?   The fastball study pitch serves as an example.   Three ballistic projectile formulas provide the answers.

     The fastball study pitch’s side view schematic shows that I released the fastball 6.05 feet in front of the pitcher’s rubber and 4.62 feet above the pitcher’s rubber with a zero degree release angle(0o).   For 2.12 feet and 0.016 seconds after release, the fastball’s initial velocity was 132.5 ft/sec.

         a.   Ballistic Projectile Time to Fall Formula

                         tF  =  V(2)(g)(syi)  +  ((vr)(sinOr))2  /  V(g)      where O indicates the Greek symbol for Theta

             1.   What Do We Know?

1.   syi  =  4.62 feet
2.   Or   =  0o
     a.   sin 0o  =  .000
     b.   cos 0o  =  1.00
     c.   tan 0o  =  .000
3.   vr  =  132.5 ft/sec
4.   g  =  32 ft/sec2

         b.   Ballistic Projectile Time to Fall Formula Calculation

1.   Time to Fall Formula      tF  =  V(2)(g)(syi)  +  ((vr)(sinOr))2  /  V(g)      where:   V indicates the square root symbol
2.   Substitute                       tF  =  V(2)(32)(4.62)  +  ((132.5)(.00))2  /  V(32)
3.   Solve the equation          tF  =  0.54 seconds

    The fastball remained in the air for 0.54 seconds.

         c.   Ballistic Projectile Total Horizontal Distance Formula

                         d  =  (vr)2(cosOr)(sinOr) +  (vr)(cosOr)V(2)(g)(syi) +  ((vr)(sinOr))2  /  (g)

             1.   What Do We Know?

1.   syi = 4.62 feet
2.   Or = 0o
     a.   sin 0o  =  .000
     b.   cos 0o  =  1.00
     c.   tan 0o  =  .000
3.   vr  =  132.5 ft/sec
4.   g  =  32 ft/sec2

         d.   Ballistic Projectile Total Horizontal Distance Calculation

1.   Total Horizontal Distance Formula    d  =  (vr)2(cosOr)(sinOr)  +  (vr)(cosOr)V(2)(g)(syi)  +  ((vr)(sinOr))2  /  (g)
2.   Substitute                                         d  =  (132.5)(1)(0)  +  (132.5)(1)V(2)(32)(4.62)  +  ((132.5)(0))2  /  (32)
3.   Solve                                               d  =  71.22 feet

     Disregarding air molecule deceleration, the fastball traveled 71.22 feet.   Air molecule deceleration would significantly decrease the fastball’s total horizontal distance.

         e.   Ballistic Projectile Trajectory Equation

                         sy  =  syi  +  tanOr(sx)  +  1/2(g)  x  (sx2)  /  ((vr)(cosOr))2

             1.   What Do We Know?

1.   syi  =  4.62 feet
2.   Or   =  0o
     a.   sin 0o  =  .000
     b.   cos 0o  =  1.00
     c.   tan 0o  =  .000
3.   vr  =  132.5 ft/sec
4.   g   =  32 ft/sec2

         f.   Ballistic Projectile Trajectory Equation Simplification

     The ballistic trajectory equation provides where airborne objects are vertically at specific horizontal displacements.   However, before substituting appropriate horizontal displacements into the equation, we simplify the trajectory equation for this fastball.

1.   Ballistic Projectile Trajectory Equation      sy  =  syi  +  tanOr(sx)  +  1/2(g)  x  (sx2)  /  ((vr)(cosOr))2
2.   Substitute                                                 sy  =  4.62  +  (0)(sx)  +  1/2(32)(sx2)  /  ((132.5)(1))2
3.   Simplify                                                    sy  =  4.62  -  (sx2)  /  1097.27

         g.   Ballistic Projectile Trajectory Equation Calculations

     Where did the fastball cross home plate?   I released the fastball 6.05 feet in front of the pitcher’s rubber.   From the pitcher’s rubber to home plate is 60.5 feet.   Therefore, the fastball traveled 54.45 feet (60.50 - 6.05) to home plate.   I prefer six horizontal points.   Dividing 54.45 feet by 6 equals aaproximately 9 feet.   Therefore, I will use multiples of 9 feet for the fastball’s horizontal displacement (sx) coordinates.

             1.   sx  =  9 feet

1.   sy  =  4.62  -  (sx2)  /  1097.27
2.   sy  =  4.62  -  (9)2  /  1097.27
3.   sy  =  4.62  -  0.07
4.   sy  =  4.55 feet

             2.   sx  =  18 feet

1.   sy  =  4.62  -  (sx2)  /  1097.27
2.   sy  =  4.62  -  (18)2  /  1097.27
3.   sy  =  4.62  -  0.30
4.   sy  =  4.32 feet

             3.   sx  =  27 feet

1.   sy  =  4.62  -  (sx2)  /  1097.27
2.   sy  =  4.62  -  (27)2  /  1097.27
3.   sy  =  4.62  -  0.66
4.   sy  =  3.96 feet

             4.   sx  =  36 feet

1.   sy  =  4.62  -  (sx2)  / 1097.27
2.   sy  =  4.62  -  (36)2  /  1097.27
3.   sy  =  4.62  -  1.18
4.   sy  =  3.44 feet

             5.   sx  =  45 feet

1.   sy  =  4.62  -  (sx2)  /  1097.27
2.   sy  =  4.62  -  (45)2  /  1097.27
3.   sy  =  4.62  -  1.85
4.   sy  =  2.77 feet

             6.   sx  =  54 feet

1.   sy  =  4.62  -  (sx2)  /  1097.27
2.   sy  =  4.62  -  (54)2  /  1097.27
3.   sy  =  4.62  -  2.66
4.   sy  =  1.96 feet 


                  Schematic 20.1:  Fastball Trajectory 

  5 -          .          .          .          .          .          .
  4 -          .          .          .          .          .          .
  3 -          .          .          .          .          .          .
  2 -          .          .          .          .          .          .
  1 -          .          .          .          .          .          .
  0 ___________.__________.__________.__________.__________.__________.
               9         18         27         36         45         54
                    Total Horizontal Distance (feet)
     With a 0o release angle and a 132.5 ft/sec release velocity, gravity alone drops the fastball 2.66 feet from a 4.62 foot height 6.05 feet in front of the pitcher’s rubber to 1.96 feet above home plate.   Therefore, gravity significantly altered the fastball’s flight path.


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