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### Over the Right Field Fence

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Date: 04/11/98 at 22:07:29
From: Richard Porter
Subject: Re: Can you help me with this?

Dear Dr. Math,

I am trying to find a formula or simply get the answer to the
following scenario:

We have a baseball field with a short right field (approx. 150 ft.)
and we want to put up a "Green Monster," (a 30 ft. high mesh net) so
it will make it more difficult to hit it out of the park. The
question is:

What would be the distance of a hit from home base over the top of the
right field fence, given the 30 ft. height?

(The obvious answer would be approximately 143 feet using the
Pythagorean Theorem; but that just doesn't seem right.)

Richard Porter
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Date: 04/12/98 at 00:17:06
From: Doctor Guy
Subject: Re: Can you help me with this?

I don't want to make this harder than it is, but I think that the
correct answer is, "That depends." What it depends on is exactly what
trajectory the ball will trace as it goes over the Green Monster.

I can imagine three different types of home runs:

(1) a high, towering home run that goes way up high (many hundreds of
feet, perhaps) and then comes down almost vertically, just on the
other side of the mesh net. This ball might only land about
155-160 feet away from home plate. This would be an easy out in
most fields.

(2) a "line drive" that goes almost in a straight line and is still on
an upward trajectory as it clears the mesh fence, and ends up
landing in a parking lot or in a field, or on the roof of a
building many hundreds of feet away. This would be a home run in
almost any field (especially in Denver, because the air is so thin
that there is less friction holding the ball back).

(3) an intermediate type of hit that follows some other type of
nearly parabolic curve, again just clearing the fence, and landing
about 180-190 feet from home plate. In most standard baseball
fields, even this would be an easy out for an infielder to catch.

Obviously these three types of hits are different: the angle that the
ball leaves the bat differs, as well as the initial velocity of the
ball, not to mention where the ball lands. I do not think it would be
fair to say that all of these balls should be said to have traveled
the same distance from home plate.

However, it may be the case that your baseball field is trying to put
a sign up on the fence to say that a ball traveled *at least* this
distance. That I can understand. If that is indeed the case, I do not
quite see how you came up with your figure of 143 feet; I think you
did sqrt(150^2 - 30^2), but you should have done sqrt(150^2 + 30^2),
which would be a bit more than 150 feet, not less; remember that the
distance from home plate to the top of the fence is like the
hypotenuse, rather than like one of the legs.

By the way, there is an excellent little book called "The Physics of
Baseball" that discusses a lot of these topics. I don't remember the
author or the publisher, but it's less than 10 years old, so almost
everything is current, and it ought to be in college libraries at
least.

Also: if you have access to a graphing calculator, and you know how to
set it into parametric mode, you can have a lot of fun modeling the
various hits that can be made to just clear the fence. You would
decide what the angle of the ball is as it leaves home plate. Let's
call this angle B, measured from the horizontal. Also, you decide what
the initial speed (velocity) of the ball is, which we'll call V. For
simplicity, we'll pretend that there is no friction from the air, and
no wind, and that the ball leaves from altitude zero. (Later, you can
fix all this.)

A little trigonometry shows how to get the vertical component of the
ball's velocity as well as the horizontal component.

oU
/|
/ |
/  |
/   |
o----o
P    H

P represents home plate. The ball starts out at velocity V,
represented by vector PU, at angle UPH from the vertical, which we
called angle B. HU is the vertical component of the ball's speed.
PH is the horizontal component of the ball's speed.

I don't know if you have studied any trigonometry or not, but here
is some:

PH/PU = cosine (B), so PH = PU * cosine (B),
which will be VOX (initial velocity in the X-direction,
i.e. horizontal)

UH/PU = sine (B), so UH = PU * sine (B),
which will be V0Y (initial velocity in the Y-direction,
i.e. vertical)

The vertical position (Y(T)) of an object in a vacuum that starts at
height H in feet, with an initial vertical velocity of V0Y feet per
second, on earth, after T seconds, is:

Y(T)= -16*T*T + V0Y*T + H

Note that V0Y and H can be positive or negative.

But, as long as we ignore friction, the horizontal position X(T) will
simply be X(T) = V0X*T, where I assume that they start at home plate,
where that is defined as zero.

On a TI 81, 82, 83, 85, or 92, you should be able to set the
calculator into parametric mode and get your calculator to graph the
path of the ball as it is "hit." You will want to set the
many baseballs stay in the air that long). Have the calculator do the
calculations of the sines and cosines for you, and store your results
in variables so you don't have so much typing to do. You can also get
the calculator to draw a line representing your Green Monster. Then
hit GRAPH and you will see if the hit goes over the wall.

Enjoy.

-Doctor Guy,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Physics/Chemistry
High School Trigonometry

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