Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Real Life Uses of Quadratic Equations

Date: 06/13/2002 at 14:08:52
From: Eric Veneto
Subject: Practical uses for quadratic equations

Can you give me a real life problem that would require you to solve a
quadratic equation? In High School, we only saw equations, not word 
problems. If we learn it, it should have practical use, right?

Date: 06/13/2002 at 17:26:45
From: Doctor Ian
Subject: Re: Practical uses for quadratic equations

Hi Eric,

Let's say I'm shooting my pistol at a watermelon 50 yards away.  
I know that the bullet comes out traveling 750 feet per second, 
and I know that the target is 150 feet away, so I can estimate 
the time of flight to be about 150 feet divided by 750 feet per 
second, or about 1/5 of a second. 

Now, I know that the bullet will drop according to 

  h = (1/2)gt^2

    = (1/2)(-32 ft/sec^2)(1/5 sec)^2

    = -16/25 feet

or a little more than half a foot.  Which means that I have to 
aim about a half a foot above the melon in order to hit it.  

But that's not really solving a quadratic equation, is it?  It's 
just taking a square root.   So let's change the problem a 
little.  I see an old abandoned refrigerator 800 yards away, and 
I want to know whether I can hit it or not. 

If I aim at a 45 degree angle, the vertical component of velocity 
will be (750 ft/sec) * sin(45 degrees), or about 530 feet per 
second.  Again, we have a parabolic arc, so the height above the 
ground after t seconds will be 

  h = (530 ft/sec)t + (1/2)(-32 ft/sec^2)t^2

Obviously the height will be zero when t=0; I want to know when 
it will be zero again at the other end!  So I set the height 
equal to zero, and I get

  0 = (530 ft/sec)t + (1/2)(-32 ft/sec^2)t^2
Now, this is still too easy, because I can just divide both 
sides by t, and then I don't have a quadratic equation anymore.  
So let's say that the refrigerator is at the top of a cliff that 
looks to be about 20 feet high.  Now I want to know when the 
height will be 20 feet:

  20 = (530 ft/sec)t + (1/2)(-32 ft/sec^2)t^2

So let's say I solve this, and find the values of t.  (There will 
be two: one for when the bullet reaches a height of 20 feet on 
the way up, and the other when it lands at the top of the cliff.)

I can multiply that by the _horizontal_ component of the velocity 
(also 530 feet per second), and if I get a distance of more than 
800 yards, then the refrigerator is in range.  If not, it's not, 
and I shouldn't waste expensive ammunition shooting at it.  

What could be more practical than that? 

- Doctor Ian, The Math Forum
Associated Topics:
High School About Math
High School Basic Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum