Super Bowl Football field Math fun problem 2/2/2008

[DBrownofdc]

too many variables

[IIIOIII]

lmao pedo

[DrewFace507943]

The equations isn't viable. The possible answers that are give are not assigned values. There for they would have to be established in order to give solvable equation. Meaning

I have no idea how tall your finger is. Nore do I know how tall you are or what type of truck is being used and its measurements.

You also don't establish a threshold of tension intended for the maximum hight that the rope can be stretched. Rephrase the question and I will give you an answer.

[dciguy01]

The answers are approximate values. The average finger height is about .75 in. The average height of a man is about 6 ft, give or take. The average height of a truck is more than twice that of a man. You don't need exact figures. Just judge based upon guesstimates. All of these values are clearly distinguishable from one another.

[TanWeiHou]

1+2=4

[Crashtian]

4.something something yards sir. D) large enough to drive a truck under.

[mathproblems]

4 something what? units?

[nlsnho]

Using the Sine law, and a bit of geometry, I calculated the height at the 50 yard line to be 7.76 yards which converts to approximately 23 feet. So my answer is D, high enough to drive a truck under it.

[mathproblems]

your answer is correct but the height is 13.43 feetand not 7.76 yds

[hjourdenjackson]

I used my cell phone to shoot my video response. I can't get the video into a single long (105 second) video on you tube, but my corrected answer (for the height at the 50 yd line) is 11 ft... small trucks can squeek under, so I still say d.

[hjourdenjackson]

Pythagoras is the right way to go, but as my good friend nemonaemo points out, you've got to work in the correct unit of measurement the whole way through.

the answer is D, big enough to drive a truck under, because in my experience, very few trucks couldn't clear 26.85 feet.

[KrelianMizrahi]

Lenght = 360 + 1 ft

at 50 yard line we have 180 ft.

sqrt(180.5^2-180^2) = 13.43 ft.

Around 4.1 meters, go go truck.

[Nemonaemo]

It's three feet to the yard, people. Most of the maths here seems to surmise that he's adding 1 yard to the rope, not the 1 foot he says in the video. As the smallest unit mentioned is foot, all maths should be done in feet. That means the length of the field is 360 feet and the length of the extended rope is 361 feet.

[Nemonaemo]

The rope forms two triagles with the ground, each with dimensions of 180ft (60y / 2), 180.5ft ((60y + 1ft) / 2) and x (desired answer).

Cosine of the smaller angle of one such triangle is 180/180.5 =~ 1

x = tan ~1 x 180 = ~3ft,

Just enough to crawl under.

[Nemonaemo]

crap, nevermind... *sigh* how often do I use trig when pythagoras suffices? Far too often...

[mathproblems]

no incorrect answer

[Nemonaemo]

Yeah, ~13.5ft. I forgot to change the cosine back into a real angle.

[mathproblems]

why are you using cosine and tan?

I still dont know how you got 13.5 ft show this please

[de1043]

1 + 1 equals 2!

[FOXTR121]

You're asking; what will be the altitude over a distance of 200ft if the rope has a length of 201ft?

Will the rope be lifted up at 1 end, or both ends?

If both, answer would be D.

If 1, it would be:

sin.(cos^-1).200/201 = altitude/201

Or:

200^2 . altitude^2 = 201^2

[FOXTR121]

Ah, you want to know the altitude if you lift it up at a point, 50 yards away from one of the ends?

Having trouble with the English language! ;)

[mathproblems]

I am sorry but I thought it was very clear in my mind anyway!!!

The answer is 13.49 feet high

[FOXTR121]

I'm not that good in English... and ft/yards are unknown in Europe!

[Nemonaemo]

Are you adding 1 yard or 1 foot of slack?

[mathproblems]

Just one foot or 1/3 of a yard is added as slack

[Yelnnek]

I apologize for the extra comment, but I realized that I had made a mistake in my calculations.

I did my work as if 1 yard had been added, not a foot.

[Yelnnek]

So, we have a right triangle with a base of 60, and a hypotenuse of 60.5.

Thus, the distance off the ground can be solved with the Pythagorean Theorem!

So, we'll have 60²+x²=60.5²

This becomes 3600+x²=3660.25

Then, x²=60.25

And sqrt(x²)=sqrt(60.25)

So, x=7.76... yards.

Plenty of space for a truck.

[mathproblems]

what is the answer in feet?

[Yelnnek]

I'd say it's high enoughw to drive a truck under, I got 7.762 yards.

I could be wrong, though.

Since the field is 120 yards long, we can cut it in half at the 50 yard line, where you'd lift it up. So, the distance from the 50-yard line a goalpost is 60 yards. We can do the same with the rope that has been lengthened a foot, and say it is 60.5 yards from the 50-yard line to the goal post. This will make a right triangle, with the distance you can lift it on the 50.

[siggy75]

D, it'll be approx 13 ft high