Kinematics motion - Dancers pausing at top of their leap

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    Kinematics Motion
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SUMMARY

The discussion focuses on the kinematics of dancers and athletes during vertical leaps, specifically the time spent in the upper half of the jump. The key equation derived is h = (gt²)/2, which describes the relationship between height (h), acceleration due to gravity (g), and time (t). Participants emphasize the importance of expressing initial velocity (vi) in terms of height and gravity to simplify calculations. The final conclusion indicates that approximately 70.7% of the total airtime is spent in the upper half of the leap.

PREREQUISITES
  • Understanding of basic kinematics principles
  • Familiarity with quadratic equations and their roots
  • Knowledge of gravitational acceleration (g = 9.81 m/s²)
  • Ability to manipulate algebraic expressions and equations
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  • Study the derivation of the equation h = (gt²)/2 in detail
  • Learn how to solve quadratic equations for time and velocity
  • Explore the concept of motion symmetry in projectile motion
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Students studying physics, particularly those interested in kinematics, athletes analyzing jump techniques, and educators teaching motion dynamics.

  • #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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