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 (0^o). 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

t_F = V(2)(g)(s_yi) + ((v_r)(sinO_r))2 / V(g)
where O indicates the Greek symbol for Theta

1. What Do We Know?

1. syi = 4.62 feet
2. Or = 0^o
a. sin 0^o = .000
b. cos 0^o = 1.00
c. tan 0^o = .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 t_F = V(2)(g)(syi) + ((vr)(sinOr))2 / V(g) where: V indicates the square root symbol
2. Substitute t_F = V(2)(32)(4.62) + ((132.5)(.00))2 / V(32)
3. Solve the equation t_F = 0.54 seconds

The fastball remained in the air for 0.54 seconds.

c. Ballistic Projectile Total Horizontal Distance Formula

d = (v_r)2(cosO_r)(sinO_r) + (v_r)(cosO_r)V(2)(g)(s_yi) + ((v_r)(sinO_r))2 / (g)

1. What Do We Know?

1. syi = 4.62 feet
2. Or = 0^o
a. sin 0^o = .000
b. cos 0^o = 1.00
c. tan 0^o = .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

s_y = s_yi + tanO_r(s_x) + 1/2(g) x (s_x²) / ((v_r)(cosO_r))2

1. What Do We Know?

1. syi = 4.62 feet
2. Or = 0^o
a. sin 0^o = .000
b. cos 0^o = 1.00
c. tan 0^o = .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) + ½(g) x (sx²) / ((vr)(cosOr))2
2. Substitute sy = 4.62 + (0)(sx) + ½(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 approximately 9 feet. Therefore, I will use multiples of 9 feet for the fastball's horizontal displacement (sx) coordinates.

1. s_x = 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. s_x = 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. s_x = 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. s_x = 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. s_x = 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. s_x = 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 5.1: Fastball Trajectory

5	.	.	.	.	.	.
4	.	.	.	.	.	.
3	.	.	.	.	.	.
2	.	.	.	.	.	.
1	.	.	.	.	.	.
0	.	.	.	.	.	.
	9	18	27	36	45	54
Total Horizontal Distance (feet)

With a 0^o 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.

Coaching Adult Pitchers