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How tall is this bridge?

  1. Oct 14, 2016 #1
    1. The problem statement, all variables and given/known data
    The question says this: Use one dimension to solve the problem.
    A woman drops a rock off a bridge (0=270 degrees-- wasn't sure how to type the line through the zero, sorry!) and it takes 2.3 seconds for the rock to hit the water below. How tall is the bridge?


    2. Relevant equations
    Delta x=Vot+one half at^2

    3. The attempt at a solution
    I have this written out:
    Delta x=Vot+one half at^2
    Delta x=(0)*(2.3 sec) plus one half times (-9.8 m/sec^2) times (2.3)^2
    Delta x= one half times (-9.8 m/sec^2) times (5.3 sec^2) ((the sec^2 is slashed out))
    Delta x=26 m
    I then made the 26 negative, as I read it needs to be.

    When I was working the problem, I wasn't sure what to do with the 0=270 degrees portion. I'd like to know if I did this correctly, and if not, what I did incorrectly/what the fix for it is. Thank you! :)
     
  2. jcsd
  3. Oct 14, 2016 #2

    haruspex

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    Neither do I. What does it mean?
    Why? It asks how tall the bridge is. That cannot be negative.
    Is that a typo?
     
  4. Oct 14, 2016 #3
    Yes, it is a typo. And I realised the negative thing a bit ago; at the time when I posted that I had misunderstood something I read, so I have 26m marked down.
     
  5. Oct 14, 2016 #4

    haruspex

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    Assuming the 270 degrees means horizontal, 26m is right.
     
  6. Oct 14, 2016 #5

    gneill

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    If you click on the ##\Sigma## icon in the edit window top tool bar, a menu of special characters and symbols will be presented from which you can select your zero with a line, otherwise know as the Greek letter theta: θ :smile:
     
  7. Oct 14, 2016 #6

    BvU

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    ##\theta=270^\circ## is more consistent with the term "dropping" in the exercise text, i.e. straight down .... :smile:
     
  8. Oct 14, 2016 #7

    haruspex

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    Yes, of course!
     
  9. Oct 14, 2016 #8
    Thanks, everyone. :)
     
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