I guess I'm no fun.  What am I missing here to make it fun enough to respond to???  Am I missing work performed to this point?

Thanks.

Thank you for the response, I'll be more patient this time.  I was unable to figure out how to attach my drawing that I'm sure would help a great deal.  Yes, I did get 127 for the adjacent of the large triangle.  I lost you at the point you stated that a new smaller triangle has been created that measures "up the slope 52'".  I came up with 54.6' up the slope as I calcualted 130' - 75'.  Like I say, I'm lost.

I'll try to make it a little more clear by giving the facts of the scenario.  This is real world - last Saturday, a motorcycle jumper jumped a claimed 310' 4" for the record.  This was measured by running a steel tape along the ground from the leading edge of the takeoff ramp to the leading edge of the landing ramp which measured 235' for the gap meaurement.  A mark was made on the landing ramp where the jumper had landed (on the 130' hypotenuse) and this was measured with a steel tape from the top of the landing ramp (28' tall oppostite side - corresponds to measurement taken along the ground from the takeoff to the landing ramps) with a distance of 75' 4" down the slope.  Since the actual distance of the jump should be measured along the adjacent only (from the leading edge of the takeoff to the corresponding point on the ground where he landed on the ramp), my argument is that the method used to measure the distance is incorrect.  I need to prove this with a formula that accurately determines the true distance travelled with the data gathered in the method that it was recorded.  Trig classes from over 20 years ago tells me they are wrong but I don't have the smarts to translate that into a formula.  He still has the record (previous record of 277.5' measured the same way) but I would like to see accurate measurements.

Thanks again,

Joe

(wish I had a nifty saying)

Outstanding.  I put it together in an Excel spreadsheet with the ratio function you suggested and it works great.  Thank you so much for your help.  On to bigger and better things. 

I am putting together a spreadsheet that will calcualte speeds for this type of jump and I am almost there.  The difficlut part is factoring in the wind resistance.  Plenty of ballistics formulas out there but none with air resistance.

Again, thank you for your persistance, much appreciated.

Thanks - Joe

Air resistance is a tough nut to crack, because it can behave in complex ways. My physics textbook gives only two:

R = bv

where R is the resistive force, a vector quantity directed opposite the direction of motion; v is the velocity of the object; and b is a constant (you know, one of those "constants" that varies in every situation) that represents the properties of the medium and shape/size of the object.

R = (1/2) DpAv²

where D is the "drag coefficient", p (should be a rho) is the density of air, A is the cross-sectional area of the object, and v is again the velocity. This formula is for objects moving at high speeds, such as planes, cars, and meteors.

Not a problem, keep bugging us. Many brains make light work.