### The Calculus of Baseball 5

```A Calculus Project By:
Matt Jaffe, Eli Greenwald, Harry Brownstein, Sarah Eisenstark and Jake Starr
In this project we explore three of
the many applications of calculus in
baseball. The physical interactions
of the game, especially the collision
of the ball and bat, are quite complex
and their models are discussed in
detail in a book by Robert Adair, The
Physics of Baseball, 3d ed. (New
York: HarperPerennial, 2002).
Question 1
It may surprise you to learn that the
collision of baseball and bat lasts only
about a thousandth of a second. Here
we calculate the average force on the
bat during this collision by first
computing the change in the ball’s
momentum.
The momentum p of an object is the
product of its mass m and its velocity
v, that is, p = mv. Suppose an object,
moving along a straight line, is acted
on by a force F = F(t) that is a
continuous function of time
Question 1, Section (a)
Prove that P(t1)- P(t0)= F(t) dt
P(t1) – P(t0) = ∫ F(t) dt
F(t) = ma(f)
mv(t1) – mv(t0) = ∫ ma(t) dt = m ∫ a(t) dt
=( ) ( )
m∫ a(t) dt = m(v(t1))-m(v(t0))
= P(t1) - P(t0)
Question 1, Section (b)
A pitcher throws a 90 mi/h fastball to the
batter, who hits a line drive directly back
to the pitcher. The ball is in contact with
the bat for 0.001 seconds and leaves the
bat with velocity of 110 mi/h. The
baseball weighs 5 oz, and in U.S.
customary units, its mass is measured in
slugs: m= w/g where g= 32 ft/s2
Question 1, Section (b), Subsection (i)
Find the change in the ball’s
momentum.
ρ = (m2v2) –(m1v1)
ρ = (49.1744 × .015)-(40.2336× .015)
ρ = .134112 kg m/s
Question 1, Section (b), Subsection (ii)
Find the average force on the bat:
J=FnetT=ΔP
ΔP= .134112
T=.001s
Fnet= 134.112N
Question 2
In this problem we calculate the work required
for a pitcher to throw a 90 mi/h fastball by first
considering kinetic energy. The Kinetic energy of
an object of mass m and velocity v is given by K=
½ mv2. Suppose an object of mass m moving in
a straight line is acted on by force F = F(s) that
depends on its position s. according to Newton's
Second Law:
F(s)= ma- m
Where a and v denote the acceleration and
velocity of the object.
Question 2, Section (a)
Show that ∫F(s) ds =½mv12- ½mv22
F(s) = ma= m(dv/dt) =mv(dv/ds)
v0= v(s0)
v1 = v(s1)
m∫ v dv
F(s) = ½mv2 from S(0) to (s1)
-
Question 2, Section (b)
How many foot-pounds of work does it
take to throw a baseball at a speed of
90 mi/h?
F= ½ Mvr2 – ½ Mv02
M = .1417
V0 = 90 = 40.233
V1 = 0
F = ½ (.1417)(0)2 – ½ (.1417)(40.233)2
F = 114.68449
F = 114.69 foot-pounds
Question 3, Section (a)
An outfielder fields a baseball 280 ft away
from home plate and throws it directly to
the catcher with an initial velocity of 100
ft/s. Assume that the velocity v(t) of the
ball after t seconds satisfies the differential
equation dv/dt = -1/10 because of air
resistance. How long does it take for the
ball to reach home plate? (Ignore any
vertical motion of the ball.)
Question 3, Section (a)
=

=

=
To anti-derivative
Ln v =
+ c  v = e-t/10 + c  100 = e0 + ec
V= e-t/10 (100)  v= -1000e-t/10+ c  d=-1000e-t/10 + c
0 = -1000e0 + c  c = 1000  280 = -1000e-t/10 + 1000
-1000
-1000
Question 3, Section (a) continued
-720 = -1000e-t/10
-1000 -1000
.72 = e-t/10
ln.72 = -t/10
-.33 = -t/10
t = 3.3s
That was the calculus of Baseball. I