Vertical Drop how to find the speed?

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Homework Help Overview

The problem involves calculating the speed of riders in free fall from a height of 42.7 m, after dropping for a distance of 21.6 m. The context is related to kinematics and the effects of gravitational acceleration on falling objects.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of time and speed during free fall, questioning the use of average speed versus instantaneous speed. There is also exploration of the correct displacement to use in calculations.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and clarifying the correct approach to finding time and speed. Some participants have offered corrections regarding the distance and the formula used, while others are verifying their understanding of the concepts involved.

Contextual Notes

There is a noted confusion regarding the correct distance to use in the calculations, as well as the significance of gravitational acceleration values. Participants are also navigating through the implications of using average versus instantaneous speed in their reasoning.

ilovemynny
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Homework Statement



The Tower of Doom at Six Flags America lifts its riders to a height of 42.7 m, then releases them to free fall for 21.6m. How fast are the riders traveling before the magnetic induction braking system starts to slow them down?

Homework Equations



t = [tex]\sqrt{}2d/g[/tex]

The Attempt at a Solution



t = (2)(21.6) / 9.81, square root
t = 4.40, square root
t = 2.10 s

21.6m / 2.10s
Speed = 10.3 m/s

I just wanted to see if I actually did this right and if I didn't can someone help me with it?

P.S. when i say ", square root" that means there's a square root in the equation
 
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No that is not right. You calculated the average speed. If the riders drop with an acceleration of 9.8m/s^2 for a certain time t, what does that tell you about their acceleration at that time t?
 
their acceleration at 2.10s x 9.81m/s/s would equal to 20.6m/s/s/s
is it wrong because there are too many seconds?
 
No, the seconds in the time will cancel one of the seconds in the acceleration. Also, upon further inspection (ok, first inspection :redface: ), you are using the wrong distance. You need to use the actual displacement, 42.7m - 21.6m = 21.1m. Also, just to be a little nit-picky, to 3 significant figures, g = 9.80m/s^2.
 
oh, wow I thought the displacement would be 21.6 because that's how long the drop down was. Huh, okay so wait should I try the same formula again but with the 21.1m?

If I did use the same formula would it be:
(2)(21.1) / 9.80 square root
t = 2.08 s?
 
Bah, sorry! I misread the problem! Yes, you're right, 21.6. I thought it said they drop TO 21.6m
 
omg i actually did do something by myself right >.<
but wait wait is the formula still right to find time?
 
ilovemynny said:
omg i actually did do something by myself right >.<
but wait wait is the formula still right to find time?

Yes, given an acceleration with no initial velocity and a displacement, d, that is how you find the time required to achieve that displacement.
 
so then time would equal to 2.1 s
and then
21.6m / 2.10s
Speed = 10.3 m/s

:biggrin: please tell me I'm right :rolleyes:
 
  • #10
No no, you had it right at post #3 except for the minor g = 9.81 or g = 9.80 thing. The velocity is just Velocity = Acceleration * Time for a constant acceleration.
 
  • #11
ohhh okay thank you so much! :!):biggrin::!):biggrin:
 

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