Math and Catapults

Date: 04/30/2003 at 23:07:44
From: Eduardo
Subject: Math and Catapults

How do you use math while operating a catapult?  

In class we are building catapults and we need to describe how we can 
use math with a catapult. (Example: think of a parabola being made 
when an object is tossed.) 

Thank you very much!

Date: 05/01/2003 at 21:57:49
From: Doctor Dotty
Subject: Re: Math and Catapults

Hi Eduardo,

Thanks for the question!

A diagram:
                          .         .             
       _ v        .          |              .
        /|    .              |                  .
       /  .                  h                      .
      /.                     |                         .
     /  x                    |                           .
    o_)______________________|____________________________.

    <------------------------d---------------------------->

V is the initial velocity, x is the angle from the horizontal that 
the projectile starts at, d is the distance moved, h is the maximum 
height attained, and t is the time taken for the complete journey.

First let's split the motion of the catapult into two directions - 
vertical and horizontal.

To do this, we first need to think of the initial velocity. We have it 
acting at a weird angle, and need to break this down into the 
horizontal and vertical components. Don't worry too much about 
these - you do them in the same way for all projectiles.

         /|
        / |
       /  |
      /   |
    v/    | 
    /     |b
   /     _|
  /)x_ _|_|
     a

The laws of trigonometry tell us that:

              b                       a
    sin(x) = ---     and    cos(x) = ---    
              v                       v

Therefore the:

  horizontal component of the initial velocity  = v cos x

  vertical component of the initial velocity  = v sin x

Let's look at the horizontal first. We have the following information:

   the time, t
   the speed, v cos x
   the distance, d

We can link them with the equation,

            distance
    speed = --------
              time

giving:

             d
  v cos x = ---
             t

This is the general equation for the horizontal motion of a 
projectile.

Now we need to think about the vertical motion.

There is a set of equations called the UVAST equations. I don't know 
whether you're familiar with them or not. If you would like to know 
where they come from, write back and I'll explain.

For any linear motion where,

   u = initial speed
   v = final speed
   a = acceleration
   s = displacement (distance from start to finish)
   t = time

The following equations are true:

     v = u + at
     s = 0.5t(u + v)
   v^2 = u^2 + 2as
     s = ut + 0.5at^2
     s = vt - 0.5at^2

So for our projectile, we have (in a vertical direction):

   u = v sin x
   [don't care about final velocity]
   a = acceleration due to gravity. On Earth, that's about 9.8. It 
       is acting downwards, so we have to treat it as negative.
   s = d
   t = t

So, using s = ut + 0.5at^2

   d = vt sin(x) + (0.5 * -9.8)t^2
   d = vt sin(x) - 4.9t^2

This is the general equation for vertical motion of a projectile on 
the earth's surface.

Let's say we have a projectile being fired from your catapult at 3m/s 
at 30 degrees from the horizontal. We can work out everything else 
about it:

                          .         .             
       _ 3        .          |              .
        /|    .              |                  .
       /  .                  h                      .
      /.                     |                         .
     /  30                   |                           .
    o_)______________________|____________________________.

    <------------------------d---------------------------->

Horizontal motion:

             d
  v cos x = ---
             t

             d
 3 cos 30 = ---
             t

             d
 2.598... = ---  (it is okay to simplify 2.598 to 3dp, as the 9.8 we
             t    will be using later on is only correct to 1dp)

   2.598t = d              (1)

Vertical motion:

   d = vt sin(x) - 4.9t^2

   d = 1.5t - 4.9t^2       (2)

So now we can substitute (1) into (2):

   2.598t = 1.5t - 4.9t^2

Then you can use the quadratic equation to solve it completely.

Using the two equations you can find out anything you need about the 
parabola, if you have enough information. You can find the maximum 
height reached by seeing when the upward velocity is equal to 0.

Does that help?  If I can help any more with this problem or any 
other, please write back!

- Doctor Dotty, The Math Forum
  http://mathforum.org/dr.math/