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### 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
http://mathforum.org/dr.math/
```
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