PHYSICS ACCORDING TO THE HORSE

OR

HOW PAINFUL CAN PHYSICS REALLY BE?

the study of a projectile



For all those wondering what this page is actually about and why it was written, this is a page describing the motion of projectiles and I made it for my physics class.  My name is Katrina, you can contact me at this address:  barnmouse@aol.com
I am going to demonstrate how a projectile works with this sample problem:
 

Lets say that there is this girl named Mo.  Mo is out in a field riding her pony Snotface.  While they are out riding they come across a large brick wall.  Now Snotface stops and stares at this wall for a while and realizes: wow, this would be a good time to buck Mo off.  "I could throw her right into that wall."  He throws her off his back three meters from the wall which is two meters tall.  The initial angle of projection is 30 degrees above the horizon.  She leaves the saddle at a height of  1.5 meters.

At what initial velocity would Snotface have to launch Mo to make her hit the wall?

How would we attack this problem?   First we would have to find the equations that describe the motion of a projectile, (Mo in this case).   A generic version of the projectile equation would be: x-x0=v0t+1/2at².  X-Xo= distance traveled.
Vo= initial velocity.  t=time.  a=acceleration.   But that equation would have to be split into the vertical and horizontal components of Mo's motion because the two components work separately.  In other words: how high Mo goes does not affect how far Mo will go, at least not directly.  Mo's horizontal and vertical movements are caused by the forces Snotface applies to Mo's body in the horizontal and vertical directions respectively.  The force Snotface applies in the horizontal direction does not affect how far Mo goes vertically and vice versa.  So, for Mo's vertical motion we will use the equation y-y0=v0yt+1/2at². Let's fill in the information we have on Mo's vertical movement. Y0, (where Mo started from) is 1.5 meters.  y, (where Mo ends up) has to be between 0 or 2 meters for Mo to hit the wall.  a is -9.8 meters per second squared, otherwise known as the acceleration due to gravity.  For Mo's horizontal motion we will use the equation x-x0=v0xt+1/2at², but the a on horizontal movement of a projectile is zero because there is no outside accelerating force. So our equation is: x-x0=v0xt.  x0 is arbitrary, we will say that it is zero. x has to be 3 meters if x0 is zero so that Mo will hit the wall.  We then substitute values into these equations and get:  0-1.5=V0yt+1/2*-9.8t²  or  2-1.5=V0yt+1/2*-9.8t²  and 3=v0xt. Then we solve the x equation for t: t=3/v0x.  Then we replace v0y and v0x with v0sin30 and v0cos30 respectively.  And we substitute the value we got for t into the y equations: -1.5=v0sin30*         3     - 9.8*           9           and
                                                                                v0cos30     2     v0²(cos30)²
.5=v0sin30*      3     -9.8*           9 .              Which simplify to  -1.5=3tan30-      44.1       and
                   v0cos30   2     v0²(cos30)²                                                    v0²(cos30)²
.5=3tan30-      44.1     .     At this point we solve both equations for v0²: v0²=             44.1              and
                v0²(cos30)²                                                                       1.5cos²30+3cos²30tan30
v0²=             -44.1           .  Then we put in approximations for the trigonometric values: v0²=       44.1        and
      .5cos²30-3tan30cos²30                                                                                       2.4240381
v0²=       -44.1      .   We then divide those out and get:  v0²=18.1927834 and v0²=47.7253048. We then take the square
        -.9240381
root of both equations, (discarding the negative square roots as not applicable), and get v0=4.3(approximate) and
v0=6.9(also approximate).   So, if you will remember that our original question was how fast Snotface would have to throw Mo off to make her hit the wall, the answer is anywhere from 4.3 meters per second, (for the bottom of the wall) to 6.9 meters per second, (to hit the top of the wall).


Now, to get back to our story, Snotface decides that he will throw Mo of at a speed of 5 meters per second. As Mo is flying through the air, her friend Fluffy sits off to the side on her horse Lazybum and tapes Mo on her handy movie camera.  Once Fluffy calls an ambulance for Mo and her many broken bones she studies her tape and records Mo's position every tenth of a second until she hits the wall:
 
      time in seconds         x-coordinate        y-coordinate
0 0             (meters) 1.5          (meters)
0.1 0.43 1.7
0.2 0.87 1.8
0.3 1.3 1.81
0.4 1.7 1.72
0.5 2.2 1.52
0.6 2.6 1.24
0.7 3 0.85

Then she takes this information and makes some graphs of Mo's motion.

  click here to view graphs


Now you might think this whole procedure is a little silly, but there are real world applications for this sort of thing besides situations like that of Snotface and Mo.  This same process could have been used to aim a cannon once upon a time, or a missile today.  Or, with a few modifications it could be used to predict where a parachute jumper leaping from a plane would land.  You could predict the movement of most any projectile.

Here are some links to sites that might tell you a little more about projectiles:
 

 PHYSICS: PROJECTILE MOTION
 PROJECTILE MOTION
 PROJECTILE MOTION & ROLLER COASTER HILLS
 PROJECTILE MOTION
 CALCULUS APOCALYPSE
 GALILEO'S WORK ON PROJECTILE MOTION

And here is a site for everyone more interested in the horse than the flight

 THE UNITED STATES DRESSAGE FEDERATION HOME PAGE

And here is a link to my teacher's web site

 MR. YOUNG'S PAGE

And that is the end of my page.
 
 

THE END