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/
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