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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum