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Projectile Motion


Date: 02/27/2001 at 22:20:45
From: D. Gilbert
Subject: Setting up quadratic equations

If a cannon is firing a projectile with an initial upward velocity of 
100 feet per second, how do you set up a quadratic equation for the 
projectile motion of the cannonball it is firing?  

I know how to solve quadratic equations, but never thought about 
setting one up.  

Thank you for your help.


Date: 02/28/2001 at 10:57:08
From: Doctor Ian
Subject: Re: Setting up quadratic equations

Hi,

Starting from first principles, we have Newton's law:

  a = F/m

We can integrate acceleration to get velocity:

  v = (F/m)t + v_i

And we can integrate velocity to get (the vertical) position:

   p = (1/2)(F/m)t^2 + (v_i)t + p_i

In this equation, (F/m) is the familiar gravitational acceleration, g; 
v_i is the initial velocity; and p_i is the initial position. 

The problem tells you v_i: it's 100 feet per second. You can choose 
p_i to be whatever you want, although it's normal to set the location 
from which the motion begins as p = 0. And g is 32 ft/sec^2 in a 
_downward_ direction.  

If you're not familiar with integration, don't worry about it.  I just 
did that to show you where the equations came from.  In most problems 
of this type, 'setting up' just means choosing the right equation, 
e.g., 

  p = (1/2)gt^2 + (v_i)t + p_i

and going from there. 

Anyway, so now you end up with 

  p = 100t - (1/2)(32)t^2 

If you choose some height above the ground, you can solve for the time 
t that it takes to reach that height. (Note that there will be two 
solutions everywhere except at the top of the arc, since what goes up 
must come back down.) Or, if you choose some time of flight, you can 
determine the height of the cannonball at that time. Note that as time 
increases, the height eventually becomes negative, which makes sense 
if you think of launching the cannonball from the edge of a cliff:

                 o
            o         o
          o             o           p > 0
 
        o                 o
      //                   
  -------                  o        p = 0


                            o       p < 0


I hope this helps.  Write back if you'd like to talk about this some 
more, or if you have any other questions. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Physics/Chemistry

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