I am constructing a model ski jump for my math and physics classes. I
need to find information about the different dimensions, physics,
math, and proportions that can be considered when constructing a ski
jump. I need information that can help me calculate the distance that
an object will go when dropped down the ramp of the jump and flung,
involving a certain speed and trajectory. Any information would be
helpful.   Thanks a lot, Winston

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Request for Question Clarification by jackburton-ga on 14 Mar 2003 01:16 PST

hi winston,
This is the most helpful piece of information i could find for you:
"How do ski jumpers use physics to get the most out of their flight?"
http://www.mansfieldct.org/schools/mms/staff/hand/lawsskijump.htm
If it answers your question, please let me know.

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Clarification of Question by winstonjones-ga on 15 Mar 2003 13:26 PST

The type of jump that i'm looking up information for is a Nordic jump,
which is uesed for jumping for distance.

A little something...

http://www.madsci.org/posts/archives/mar98/891200231.Ph.r.html

When the skier leaves the jump gravitational potential energy will
have been converted into kinetic energy.  The difference in height
between the top of the ski jump and the end of the ski jump times
gravity times mass will give you the energy.  Then you will be able to
calculate the velcoity of the skier as they leave the jump.

The skier will leave the ramp at a particular angle (a) to the
horizontal.  If there velocity is v.  Then v(horiz) = v * cos (a) and
v(vertical) = v * sin(a).
v(vertical) will determine how long they spend in the air. 
v(vertical) combined with v(horiztonal) will determine how far they
travel.

Now the skier will rise a v(vertical) is converted into gravitational
potential energy again.  Now at some point the skier will reach a
maximum height and then start dropping.

Where they land depends on what sort of surface they land on.
If it is flat then the caluclation is simple.
If it is a straight slope then the calculation is a bit harder.
If it is a curved slope then the calculation is harder still.

Of course this doesn't take into account drag and friction.

Since none of the paid researchers have provided an answer yet, I
thought I would add some further comments to my last comment.

The flight path of the skier when they leave the ramp will be a
parabola.  It is easy to express this as parametric equations relative
to time.
The x positon is easy to determine
x(t) = v(horiz) * t

The y position is harder to determine
y(t) = a * t^2 + b * t + c
Where a, b and c are unknown constants which can be worked out as
follows.

When t = 0 then we will declare the end of the ramp is at position
(0,0) so...
y(0) = a * 0 + b * 0 + c = 0
c = 0

The derivative of y(t), y'(t) = 1/2 * a * t + b
y'(t) is the vertical speed, we know that the speed at time 0 =
v(vert) so...
y'(0) = 1/2 * a * 0 + b = v(vert)
b = v(vert)

THe second derivative of y(t), y''(t) = 1/2 a
y''(t) is the vertical acceleration, which in this case we know what
it is as it is gravity.
y''(t) = 1/2 * a = 9.8 m/s/s

Now we have two equations (x(t) and y(t))which tell us where the skier
is inrespect to time after they leave the ramp.  These can be combined
by subsituting t = x(t) / v(horiz) into y(t) to give us an equation
that gives y in terms of x.

Now we have to declare a function that represents the slope of the
mountain.  In one of the other comments there is a link to a page
saying that the slope of the mountains are parabolas.
y = p * x ^2 + q * x + r
Where p, q and r are again fixed constants.

Now you have two equations the first gives the flight of the skier and
the second gives the shape of the slope.  Where the curves of these
equations meet the skier lands on the slope.

So it is simply a matter of sloving the set of equations and you will
find the position x where the skier lands.  You will also be able to
work out how long he was in flight.