Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**First Guess****Member**- Registered: 2005-11-03
- Posts: 5

I can't quite figure out the formula required to determine the true adjacent side distance that corresponds to a measured distance down the slope of the hypotenuse when the overall opposite and hypotenuse sides lengths are known on a right triangle. The opposite side is 28 feet tall, the hypotenuse is 130 feet long and the measured distance down the slope of the hypotenuse is 75 feet. How do I derive the corresponding distance along the adjacent axis?

I managed to figure out the overall adjacent length with the known opposite and hypotenuse lengths but can't get that last part.

Any help at all would be most appreciated. Thanks in advance.

Offline

**First Guess****Member**- Registered: 2005-11-03
- Posts: 5

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.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Give a day or so for someone to respond ...

So, you have solved the right triangle using Pythagoras Theorem **a²+b²=c²**, but you now want another measurement, correct?

BTW A sketch would be handy - you can upload it with your message (press Post reply and look for the upload slots selection)

But I think I can understand what you want ...

First let's solve the big triangle:

a²+b²=c² =⇒ a = √(130² - 28² ) = √(130² - 28² ) = 127

Is that what you got?

Now, if you measure down the hypotenuse 75' you would be 127-75 = 52' up the slope.

Now you have a new (smaller) triangle that has an adjacent of 52', and the same angle in the corner, so you could just ratio off the lengths.

So, the smaller triangles hyptenuse would be (52/127) * 130 = 53.2

And the smaller triangles opposite would be (52/127) * 28 = 11.5

Or then again, I may have got this all mixed up

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**First Guess****Member**- Registered: 2005-11-03
- Posts: 5

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)

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

If you mean the horizontal distance is the only one that counts, then that can be calculated by ratio.

You know the ramp's slope measures 130', but along the ground it is only 127', so all slope measurements should be ratiod accordingly.

So the 75' 4'' (75.333') becomes 75.333 x (127/130) = 73.59' (73' 7'')

Add that to the 235' and the total horizontal jump = 235 + 73.59 = 308.59'

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**First Guess****Member**- Registered: 2005-11-03
- Posts: 5

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

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Coool ... I am glad I was able to help!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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) D*p*Av²

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.

El que pega primero pega dos veces.

Offline

**First Guess****Member**- Registered: 2005-11-03
- Posts: 5

Thank you for the input. I have seen the R=bv equation but I have not seen the other. That one looks the most promising to date. The v portion will prove to be difficult as a secondary function will need to be performed that needs to factor in the increasing vertical and diminishing horizontal speeds (the bike is powerless in midair) and also to include gravity with all of that to be continually updated as the bike travels through it's arc. To add to the difficulty, the slope of the landing ramp needs to be factored in as well.

The good thing about it is I have real data to bounce off of the speed calculator as a data checker. Distances of 30 feet to 300 feet, takeoff angles of 11, 16, 22 and 28 degrees, landing angles of 8, 12 and 14 degrees and speeds from 33 to 82 mph.

Again, thanks for the input. I guess my next step is to take a class or two and figure these things out for myself so I don't have to keep bugging you folks.

Joe.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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

Offline