Optimal Acceleration for Long Jump on the Moon

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Homework Help Overview

The discussion revolves around determining the maximum rate of freefall acceleration on the moon necessary for a long jump of 7.52 m, with a maximum running speed of 5.90 m/s. The context involves projectile motion and the specific conditions of jumping on the moon.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of projectile motion equations, particularly focusing on the angle for maximum range and the implications of the sine function in the context of the problem. Questions about angle measurements and their relevance to the maximum range are raised.

Discussion Status

The discussion is active, with participants exploring different aspects of the problem, including the significance of the angle for maximum range and the interpretation of the sine function. Some guidance has been offered regarding the angle of 45 degrees, but there is still uncertainty about the implications of this angle on the calculations.

Contextual Notes

Participants express confusion regarding the angle measurement and the conditions under which the maximum range is achieved. There is also a mention of the need for clarity in indicating logic when solving problems, reflecting the challenges faced in understanding the mathematical relationships involved.

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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