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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/2
3. sf = 10,000 feet
4. si = 0 feet
c. Falling
Baseball Final Velocity Calculation
1.
vf = V(vi2) + 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)(RV2)
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)(RV2)
2. Substitute Wt = (AD)(SF)(CSA)(MFV2)
3. Divide both sides by (AD)(SF)(CSA) Wt / (AD)(SF)(CSA) = (AD)(SF)(CSA)(MFV2)
/ (AD)(SF)(CSA)
4. Because (AD)(SF)(CSA)/(AD)(SF)CSA) = 1 Wt / (AD)(SF)(CSA)
= MFV2
5. Change sides MFV2 = Wt / (AD)(SF)(CSA)
6. Take the square root of both sides V(MFV2) = 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/sec2 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. |