# Help with question relating to long jumps! Unsolvable by math teacher!

1. Oct 4, 2011

1. The problem statement, all variables and given/known data
This problem was given as extra credit to our class. My teacher was unable to solve this problem. She had asked several math teachers as well as both my parents who are electricalmengineers and commonly use math. This is the exact wording of the problem.

In the 1964 Olympics in Tokyo, the best men’s high jump was 2.18 m. *Four years later in Mexico City, the gold medal in the same event was for a jump of 2.24 m. *Because of Mexico City’s altitude (2400 m), the acceleration of gravity there is lower than that in Tokyo by about 0.01 m/s2. *Suppose a high-jumper has a mass of 72 kg:
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a. Compare his mass and weight in the two locations
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b. Assume that he is able to jump with the same initial vertical velocity in both locations, and that all other conditions are the same except for gravity. *How much higher should he be able to jump in Mexico City?

2. Relevant equations

3. The attempt at a solution

2. Oct 4, 2011

### Staff: Mentor

So, the problem is that your teachers are not very good?

You have to show some work on the problem before we can help (Forum rules). The question involves basic kinematics, and is not too difficult.

3. Oct 4, 2011

### Johnson

I agree with gneill, this is a very basic question. I don't understand how high school math or physics teachers can't get this, or even your parents if theyre engineers.

Attempt it and I'll give you a hand if you need it.

4. Oct 5, 2011

### PhanthomJay

In fairness to your (high school?) math teachers and parents, they would have to know the kinematic equations of motion and the value of the acceleration of Gravity in Tokyo to 3 significant figures, in the SI system of measure, in order to solve this problem.

5. Oct 5, 2011

### Staff: Mentor

Hmph. The difference in elevation is given, so a reasonable estimate of the change in local g is possible. The extra information required is a rough value for the radius of the Earth.

This will make it possible to compare the magnitude of the change in high jumps as actually measured with that which might be due to location change.

6. Oct 5, 2011

### cepheid

Staff Emeritus
I agree with gneill. Without worrying about absolute numbers, it should be easy enough to get the relative difference in weight between the two locations, and from that, the difference in height achieved.

EDIT, well not exactly. You can get the ratio of heights in terms of the ratio of g's. For an actual height difference, then, you'll need to know g in one of the locations.

Last edited: Oct 5, 2011
7. Oct 5, 2011

### D H

Staff Emeritus
You don't need to know g in one of the locations. You do need to know that the difference between the gravitational accelerations $\Delta g=g_t-g_m$ is small, and therefore that

$$\frac{g_t}{g_m} = \frac{g_t}{g_t-\Delta g} = \frac{1}{1-\Delta g/g_t} \approx 1+\Delta g/g_t$$

This delta g is small. Using a slightly incorrect version of the gravitational acceleration in Tokyo, for example, 9.80665 m/s2 instead of the correct value of 9.798 will result in a negligible error. Even if you use a crude estimate of 10 m/s2 you will still get pretty much the same answer.

Besides, the error will be much larger if you use the stated difference in gravitational acceleration between Tokyo and Mexico City. The correct value is 9.798 m/s2 - 9.779 m/s2 = 0.019 m/s2, not 0.01.

Which leads me to a rant.
This is one of those terrible and misleading physics problems that gives the wrong information, beg sfor an incorrect solution, and leaves out some incredibly important pieces of information.

Re the wrong information: The acceleration of gravity is about 0.019 m/s2 lower in Mexico City than in Tokyo. This would lead to an answer that is off by a factor of 2 were it not for a couple of other problems with this problem.

Re begging for an incorrect solution: The problem implies that these high jumpers jump over 2 meters. From a physics perspective, they jump a bit over a meter. The pole is over 2 meters off the ground. Jumping 2+ meters would require those high jumpers to start from a prone position. They start the jump with their center of mass at about 0.57 times their height already situated above the ground. Here the factor of two error in Δg will pretty much offset the factor of two error in the height of the jump. If you happen to get the right answer on a test question by means of two offsetting errors and you might get a point for sheer dumb luck. You will not get full credit.

Re the incredibly important pieces of information that were completely left out:
1. Valery Brumel was 10 cm shorter than Dick Fosbury. The difference in the heights of their centers of mass alone fully explains the 6 cm difference in the heights of their jumps.
2. In fact, Fosbury most likely jumped several centimeters less than did Brumel. Brumel used the straddle technique, which requires the center of mass to rise above the height of the bar. Fosbury used the Fosbury flop. One huge advantage of the flop is that the center of mass can stay below the bar at all times. Nowadays a jumper can clear the bar with his center of mass twenty centimeter below the bar. While Fosbury hadn't quite perfected the technique in 1968 to the extent that modern jumpers have, his center of mass was most likely at or below bar level.

Gravity has nothing (well, almost nothing) to do with the six centimeter difference in height between the two jumps. Gravity accounts for a couple of millimeters of that difference. That "all other conditions are the same" is completely incorrect is what explains the difference in the heights of the two jumps.

8. Oct 5, 2011

### Staff: Mentor

Great post, D H.