Over the Right Field Fence

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.)

Thanks for your consideration,

Richard Porter

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 
x-coordinates from about -10 to about 250, and the y-coordinates from 
about -10 to about 200, and have T go from about 0 to 15. (Not too 
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/