1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Where does the arrow land?

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data
    An arrow is fired horizontally off a cliff. The initial speed of the arrow is 250 ft/s and the cliff is 22 ft high. How far from the base of the cliff does the arrow land?


    2. Relevant equations

    y=1/2a(t^2)
    x=vt
    3. The attempt at a solution

    Vertical:

    y=1/2 a(t^2)
    t=(√2y)/a

    t=2.12 s

    Horizontal:
    x=vt
    x=250(2.12)
    x=530 ft.

    Is this answer correct? Thanks in advance! :smile:
     
  2. jcsd
  3. Dec 29, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    The problem distances are given in feet, and I think you are using g = 9.8 m/s^2. I hope you realize that feet and meters can't be used interchangeably.
     
  4. Dec 29, 2013 #3
    I do realize that, I just forgot to convert. I can't find anywhere where the units clash, though, so is it right anyway?

    Thanks again! :smile:
     
  5. Dec 29, 2013 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If you have height in feet and a (or g) in m/s^2, I would say the units "clash".
     
  6. Dec 30, 2013 #5
    Ah, so would I. I didn't see that at first. But I suppose I phrased my question wrong. Did it affect the answer numerically? Is the time still right, and I just got the units wrong? Or do I need to convert for some reason?

    Thanks again!
     
  7. Dec 30, 2013 #6

    adjacent

    User Avatar
    Gold Member

    First learn to convert all the units in to SI units.Then you will be fine.(If you do it correctly)
     
  8. Dec 30, 2013 #7
    Sorry, I must be phrasing something wrong. I understand that I neglected to convert the units. I know how, I just forgot. What I'm really trying to ask here is if, in this particular case, it affected the answer numerically. I got the units wrong, but was the number part right? I'm failing to see where the units would have affected my numerical answer.

    Thanks!
     
  9. Dec 30, 2013 #8

    adjacent

    User Avatar
    Gold Member

    What did you use for acceleration? 9.8m/s2 ?
    or did you convert that to feet?
    What is the difference between 9.8meter/s2 and 9.8feet/s2 ?
    It all depends on what you used for a
     
  10. Dec 30, 2013 #9
    I used meters. I think I see it now! Was the time wrong? I see that I used feet for y and meters for a, is that what is wrong with the answer?
     
  11. Dec 30, 2013 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    9.81 m is not the same as 9.81 ft, is it? What length is one ft in meters?

    When you use an equation, the units are together with the data. So you got the numerical result for time [tex]t=\sqrt{\frac{2y}{a}}[/tex] as

    [tex]t=\sqrt{\frac{44 ft} {9.81ms^{-2 }}}=2.12 s \sqrt{\frac{ft}{m}}[/tex], in [s sqrt(ft/m)] units.
    If you multiply it with the speed in ft/s, you get the distance in [itex]\sqrt{\frac{ft^3}{m}}[/itex].

    Does it have sense?

    ehild
     
  12. Dec 30, 2013 #11

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes, units are not a style choice: they do affect the numerical values in calculations.

    Hint: since you are working in feet, g is taken as 32.2 ft/s^2. You don't necessarily have to convert all your calculations if you don't want to.
     
  13. Dec 30, 2013 #12

    adjacent

    User Avatar
    Gold Member

    Lets see now.
    You used the equation:y=1/2a(t^2)
    So ##t=\sqrt{\frac{2y}{a}}## (ah your equation was wrong.You wrote ##t=\frac{\sqrt{2}y}{a}##
    If a=9.8m/s^2,
    ##t=\sqrt{\frac{2*22}{9.8}}##
    =2.12s
    We used feet for y and meters for a.This is wrong,right?
    Then if we use feet instead of meters for a,
    ##t=\sqrt{\frac{2*22}{32.1522}}##
    =1.17s
    This answer is right because we used feet for both y and a.
    If we use meters for both y and a,
    ##t=\sqrt{\frac{2*6.7056}{9.8}}##
    =1.17s

    The last two answers were same.
    because we used the same units for y and a.
     
  14. Dec 30, 2013 #13
    Ehelid and SteamKing,thanks for the explanations. I know units aren't optional, I just didn't see *where* they came into play for this particular problem, so I was trying to achieve some level of clarification on that before moving on.

    Thank you, adjacent, that was an incredibly helpful explanation. That's what I thought, I just didn't think about the "y" being in feet at first. Thanks for clarifying where I went wrong!

    Okay, I know you already calculated the time, adjacent, but I went over it again anyway just to be sure I understood. Could someone please tell me where the square root button can be found?

    I, too, got 1.17 seconds for the time.

    Horizontal:
    x=vt
    x=250ft(1.17s)
    x=293 ft

    How's that?
     
  15. Dec 30, 2013 #14

    adjacent

    User Avatar
    Gold Member

    It can be found on the right side of the post of the Advanced editor.(Quick symbols)
    You can also use LaTeX. At the top right side of the advanced editor


    That's right
     
  16. Dec 30, 2013 #15
    Ah, found it! Thank you. That's it for this problem, correct? Thanks again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Where does the arrow land?
Loading...