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 1It 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

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 (a)

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)

Find the change in the ball’s momentum.

ρ = (m2v2) –(m1v1)

ρ = (49.1744 × .015)-(40.2336× .015)

Question 1, Section (b), Subsection (i)

ρ = .134112 kg m/s

Find the average force on the bat:J=FnetT=ΔP

ΔP= .134112T=.001s

Question 1, Section (b), Subsection (ii)

Fnet= 134.112N

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- mWhere a and v denote the acceleration and velocity of the object.

Question 2

Show that ∫F(s) ds =½mv12- ½mv2

2

F(s) = ma= m(dv/dt) =mv(dv/ds) v0= v(s0)

v1 = v(s1)

m∫ v dvF(s) = ½mv2 from S(0) to (s1)

Question 2, Section (a)

-

How many foot-pounds of work does it take to throw a baseball at a speed of 90 mi/h?

F= ½ Mvr2 – ½ Mv0

2

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 2, Section (b)

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)

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 + c0 = -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 hope that you all think about this

when you watch baseball from now on.

The Indians are the best team in baseball as of now….

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