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How high is the cliff and speed of arrow?

  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 280 ft/s and the arrow lands 400 feet from the base of the cliff. How high is the cliff?

    2. Relevant equations
    t=x/v
    y=1/2(a)(t^2)

    3. The attempt at a solution

    I did the horizontal component first.

    xinitial=400
    a=0
    vinital=280 ft/s
    x=vt
    t=x/v
    t=400/280
    t=1.43 s

    Vertical:
    t=1.43 s
    a=9.8 m/s^2
    y0=0
    y=?

    y=1/2(a)(t^2)
    y=1/2(9.8)(1.43)^2
    y=9.8(2.04)^2

    y=20 ft tall.

    Is this correct? If so, why didn't I have to half my answer?

    Thanks in advance! :smile:
     
  2. jcsd
  3. Dec 29, 2013 #2

    SteamKing

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    Your algebra is a little fuzzy when you are calculating y. Numbers seem to appear and disappear without reason.

    Also, the problem distances are given in feet, and 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'm not sure what you mean by "numbers". Would you mind elaborating which ones? Part of it may have been done mentally.
    Yes, I realize that feet and meters can't be used interchangeably, I just got used to using meters and forgot to convert. However, is converting essential for this problem? I don't see a step where the units clash, but I could just be looking over one. It looks like I have feet with feet and meters with meters throughout the problem, but again, I could be wrong. Same for the other problem. Are my answers incorrect, or were you just pointing out areas where you thought I may be unclear? Thanks again!
     
  5. Dec 29, 2013 #4

    SteamKing

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    Specifically:

    y=1/2(a)(t^2)
    y=1/2(9.8)(1.43)^2 - I understand this, since you have previously solved for t

    But a = 9.8 m/s^2, and if you multiply 9.8 m/s^2 by time squared, your remaining units will be meters, not feet.


    y=9.8(2.04)^2 - where does 2.04 come from? What happened to the (1/2) and t = 1.43 s?

    y=20 ft tall. - Again, how did you arrive at this answer from the previous step?
     
  6. Dec 30, 2013 #5
    Ah, I see. Okay, the 2.04 came from squaring the 1.43. I just left the ^2 on, oops. I multiplied the 9.8 by the 2.04 to get 19.992, and rounded to 20. The units should have been meters. Do I half the answer? I thought so, but in an example problem my teacher did similar to this, he didn't half the answer, so I thought I may have been missing something cancel out.

    Thanks again!
     
  7. Dec 30, 2013 #6

    PhanthomJay

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    You cannot just dump the 1/2 ....it is part of the equation. Also, please determine the height of the cliff in feet! You may use g = 32.2. ft/sec^2.
     
  8. Dec 30, 2013 #7

    adjacent

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    Medgirl314,
    In your earlier thread,Please read the posts on why you can't use feet and meters interchangeably.
     
  9. Dec 30, 2013 #8
    PhantomJay, I didn't think I could just dump the 1/2, but in a previous problem, it seemed that my teacher did, so I thought maybe it canceled out somewhere. I see it doesn't now, so I can fix the problem. Adjacent, I know you can't use feet and meters interchangeably, but I just forgot for these problems and was trying to find where the units imposed a problem. I see where now, so I'll come back to this problem after we finish the other one.

    Thanks! :smile:
     
  10. Dec 30, 2013 #9
    Is the answer 32.9 ft. ?
     
  11. Dec 30, 2013 #10

    adjacent

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    Right :smile:
     
  12. Dec 30, 2013 #11
    Thanks! :D I should probably round to 33 ft for this problem. Thanks again!
     
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