Kinematics motion - Dancers pausing at top of their leap

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The discussion focuses on the kinematics of dancers and athletes appearing to "hang" at the peak of their jumps. Participants analyze the time spent in the upper half of a leap, emphasizing the need to express time as a function of height (h) and gravitational acceleration (g) without initial velocity (vi). The conversation includes solving quadratic equations and clarifying the correct use of kinematic equations, particularly regarding the definitions of initial and final velocities. It is noted that dancers may exaggerate their movements to enhance this "hanging" illusion, impacting their overall height. Ultimately, the correct ratio of time spent in the upper half of the jump is determined to be approximately 70.7%.
  • #31
TSny said:
Seems like that question is leading us aside. We have a couple of choices. We can continue your line of thinking (which is getting you pretty close to the result) or we can step back and start all over with a different approach that will lead to the answer with the least amount of calculation (for example, haruspex's method).

Which would you like to do?

I'll like to finish up with my line of reasoning.
Capture3.JPG


Then run another loop using a different method.
 
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  • #32
OK, it's getting pretty late here where I am. But let's finish up your line of reasoning.

So, going back to ##h-h/2=.5(v_i+v_f)t##, what do you get for t after using your result for ##v_i##?

[EDIT. Oops , I see you already have done this. But you made a mistake in the algebra. h should appear in your answer for t.]
 
  • #33
TSny said:
OK, it's getting pretty late here where I am. But let's finish up your line of reasoning.

So, going back to ##h-h/2=.5(v_i+v_f)t##, what do you get for t after using your result for ##v_i##?

As above

If the above is true, then, what follows is basic division and algebraic manipulation.
 
  • #34
Your answer for t should contain h as well as g.
 
  • #35
TSny said:
Your answer for t should contain h as well as g.

It did until I did some cancelling
 
  • #36
Show the steps you used so we can identify your mistake.
 
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  • #37
tsny said:
show the steps you used so we can identify your mistake.

Capture3.JPG
 
  • #38
Good until the last line. How would you simplify ##\frac{h}{\sqrt{h}}##?
 
  • #39
TSny said:
Good until the last line. How would you simplify ##\frac{h}{\sqrt{h}}##?

h.h^-0.5
 
Last edited:
  • #40
TSny said:
Good until the last line. How would you simplify ##\frac{h}{\sqrt{h}}##?

above
 
  • #41
Can you simplify ##h\cdot h^{-1/2}##?
 
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  • #42
TSny said:
Can you simplify ##h\;h^{-1/2}##?

equals h^(0.5)
sqrt(h)

This is it.
 
  • #43
Good. So you have shown
##\frac{h}{\sqrt{-gh}} = \frac{h}{\sqrt{-g}\;\sqrt{h}} = \frac{h}{\sqrt{-g}\;h^{1/2}} = \frac{\sqrt{h}}{\sqrt{-g}}= \sqrt{-\frac{h}{g}} ##
 
  • #44
So, you have now found the time to go from 0 to h and also the time to go from h/2 to h.
 
  • #45
TSny said:
So, you have now found the time to go from 0 to h and also the time to go from h/2 to h.

So it's just basic division and cancellation left.
 
  • #46
Yes.
 
  • #47
TSny said:
Yes.

I'll finish it up and post it. Thanks!
 
  • #48
OK. I will check tomorrow. Cheers!
 
  • #49
Answer is 70.7%
 
  • #50
Yes, that's the right answer.
 

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