Optimal Acceleration for Long Jump on the Moon

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SUMMARY

The discussion focuses on calculating the maximum rate of freefall acceleration required on the moon for an athlete to achieve a long jump of 7.52 meters, surpassing Galina Chistyakova's record. The maximum running speed is given as 5.90 m/s, and the analysis involves projectile motion equations. The angle for maximum range is determined to be 45 degrees, leading to a calculated freefall acceleration of -4.63 m/s². The sine function's properties are discussed to clarify the conditions for achieving maximum range in projectile motion.

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kristen151027
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I have what probably sounds like a simple question...here it is:

You desperately want to qualify for the Olympics in the long jump, so you decide to hold the qualifying event on the moon of your choice. You need to jump 7.52 m (and conveniently beat Galina Chistyakova's record) to qualify. The maximum speed at which you can run at any location is 5.90 m/s. What is the magnitude of the maximum rate of freefall acceleration the moon can have for you to achieve your dream?

equations:
∆x=-v^2sin2Φ/a_y
Φ = 0.5arcsin [-a_y∆x/v^2]
there are other equations but I don't know which ones to use
 
Last edited:
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Do you know projectile motion analysis ?

Hint: Use the equation for maximum range .
 
We've been studying projectiles, but nothing too indepth. I tried to use the first equation I listed (the maximum range one...?), but I ran into trouble with the angle measurement.
 
At what angle does the projectile or jumper attain maximum range ?
 
No idea...is that something the question should provide?
 
In your first expression, ∆x becomes max. when sin2Φ = 1.
Therefore Φ equals ___ ?
 
Φ = 45 degrees
(because sin2Φ = 2sinΦcosΦ
and when Φ = 45 ... it's 2*(1/root 2)*(1/root 2)...which is 1)
if Φ = 45 degrees, then the a_y = -4.63 m/s^2
therefore, the magnitude is 4.63
correct?
 
One thing I'm not quite clear on is how "∆x becomes max. when sin2Φ = 1" ...I'm probably just not thinking clearly about it. The answer is asking for the magnitude, not the direction, so the sign of the answer doesn't matter.
 
Yes, you are right :smile:
Edit: What are the values that sine function can take ?
 
  • #10
The sine function can take values of 0 to 1. Ah and it can't be 0, so one is the maximum, but when sinx = 1, cosx = 0. Therefore, it has to be somewhere in between. Now the question is how to indicate such logic concisely when doing a problem. (I very much appreciate your help, by the way!)

Edit: sine can take values of -1 to 1...oops
 
Last edited:
  • #11
Just got it. When Φ = 45 degrees, that's creates the maximum value because we're dealing with a double-angle. So that explains why the answer is highest when Φ = 45 degrees.

Thanks again for your help! :biggrin:
 
  • #12
Correct and you're welcome :smile:
 

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