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Free-Fall Acceleration Problem

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A bolt is dropped from a bridge under construction, falling 85 m to the valley below the bridge. (a) In how much time does it pass through the last 28% of its fall? What is its speed (b) when it begins that last 28% of its fall and (c) when it reaches the valley beneath the bridge?


    2. Relevant equations
    No idea.


    3. The attempt at a solution
    I did .28 x 85 = 23.8 m. I also assumed g = 9.8 m/s^2, xinitial = 23.8, xfinal = 0, but I'm now lost. The five constant acceleration constants need v or t which I have neither.
     
  2. jcsd
  3. Sep 8, 2009 #2
    does the question require you to take air resistance into account? Or should I assume that it's dropped in vacuum?

    I won't tell you how to solve it, but I'll just give you a hint and some relavent equations. (As stated by the forum rules, we don't do your homework for you.)

    Hint: Draw a speed-time graph. Assign variables to the unknown numbers.

    Relavent equations:

    a = (v-u)/t
    v = d/t

    Good thing to remember: Always draw the speed-time graph or velocity-time graph if you're unsure on how to solve a kinematics question. (just a rough sketch of it will do, you don't have to draw it accurately on graph paper)
     
    Last edited: Sep 8, 2009
  4. Sep 9, 2009 #3
    I believe I'm suppose to neglect air resistance. How could I draw a speed-time graph, if I don't have velocity or time? And what is "u" in the a = (v-u)/t equation?
     
  5. Sep 9, 2009 #4
    Well, first you might want to figure out the information for the first 72% of the fall - that will help you get the starting velocity when the bolt enters the final 28%, and from there you can find all the other information you need.
     
  6. Sep 9, 2009 #5
    Would a = -9.8 or +9.8?

    Okay I did part b without doing part a if that is possible. First I flipped the problem upside down so the starting point is 0m and the bolt is falling up to 85m so that a = 9.8. I did v2 = vo2 + 2a(x-xo) to find v. So it's v2=0 + 2(9.8)(61.2-0). So v = 24.5m/s. That's part b.

    But I'm confused on what part a is asking. Does it mean how much time elapsed during the last 28% of the fall or in my case the flight upward?
     
  7. Sep 9, 2009 #6
    Acceleration is technically -9.8 because it's in a downward direction, but your method looks sound to me. I mean, the problem is asking for speed anyway; therefore, you don't need the negative sign to indicate direction, so you're fine - but just be aware that if you come across something similar asking for velocity, it would be negative.

    And yes, part a is asking how much time elapses during the final 28% of the fall.
     
  8. Sep 9, 2009 #7
    For part a I did x-xo=vot + 1/2at2. Plugging in I get

    85-61.2=24.5t + 1/2(9.8)t2 to get 4.9t2 +24.5t - 23.8 = 0

    Okay, that's nice, but when I tried doing quadratic formula, both my times come out to be negative. t = -1.32s and -3.68. What did I do wrong?
     
  9. Sep 9, 2009 #8
    I'm thinking maybe you made an error with the quadratic formula.
     
  10. Sep 9, 2009 #9
    My god, you are right, I made the answer to -4ac to negative. Ok so then the answer for a must be .8seconds. Does that sound too small? Also I found the answer for part c to be 33 m/s.
     
  11. Sep 9, 2009 #10
    No, that all sounds about right. :)
     
  12. Sep 9, 2009 #11
    Thanks so much physicsface! Great help!
     
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