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Boulder falls from cliff -- how high is the cliff?

  1. Nov 9, 2016 #1
    1. The problem statement, all variables and given/known data
    I searched around and found a manual, so this time I have the complete question in English (I'll use different numbers though, because mine is the 8th edition, and the manual is from the 14th, but it's not something that affects this exercise):

    2.98) An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.50 s for the boulder to fall the last third of the way to the ground. You may ignore air resistance. (a) What is the height of the cliff in meters? (b) If in part (a) you get two solutions of a quadratic equation and you use one for your answer, what does the other solution represent?

    2. Relevant equations
    y = y0 +U0*t + 1/2 * a * t^2 (1)

    U = U0 + a * t (2)

    3. The attempt at a solution
    The assumption I made/parameters I took:

    ----------------| ^ * Positive Direction is upwards.
    | | (+) * y = 0 is at the bottom of the cliff/the ground.
    | * H = 3h
    | * H is the whole height of the cliff.
    | * | a | = | g | = 9,8 m/s^2
    |
    |
    |
    |
    |
    |
    -------------- y = 0


    - y0 = H:

    Relevant Notes:

    - t(2H/3) is the time I set it takes for the rock to drop from heigh H to h. It's not a funtction, or anything like that.

    - U(H/3) is the speed I set that the rock has when it arrives at heigh h.

    H = 3h -> h: (1) => h = 3h - 4,9 * m/s^2 * t(2H/3)^2 <=> t(2H/3)^2 = 0,4 * s^2/m * h <=>

    <=> t(2h/3) = 0,63 * s/sqrt(m) * sqrt(h)

    (2) => U(2H/3) = -4,9 * m/s^ * t(2H/3)

    -y0 = h:

    Relevant Notes:

    - U0 in this case is U(2H/3.

    -t(h) is give to us by the book, which is 1,50 s.

    h -> 0: (1) => 0 = h + U(2H/3) * t(h) + 1/2 * a * t(h)^2 <=>

    <=> 0 = h - 4,9 * m/s^2 * t(2H/3) * 1,50 s - 4,9 * m/s^2 * 2,25 * s^2 <=>

    <=> h = 7,35 * m/s * t(2H/3) + 11,025 * m <=> h = 7,35 * m/s * [0,63 * s/sqrt(m) * sqrt(h)] + 11,025 * m

    -Now, I set sqrt(h) = u, therefore h = u^2

    -Also, I set sqrt(m) = n, so m = n^2


    From where we left of:

    u^2 = 4,63 * u * n^2/s * s/n + 11,025 * n^2 <=> u^2 - 4,63 * n * u -11,025 * n^2 = 0

    Δ = β^2 - 4αγ = (-4,63 * n)^2 - 4 * (1) * (-11,025 * n^2) = 21, 43 * n62 + 44,1 * n^2 = 65,53 * n^2

    u 1,2 = (+ 4,63 * n +/- sqrt(Δ))/2 = (4,63 +/- 8,09) * n/2 <=> u1 = 6,36 * n & u2 = -1,73 * n

    Naturally, u1 is the correct answer, but u & n are variables I set myself. So:

    sqrt(h) = 6,36 * sqrt(m)

    I square both sides of the equation: h = 40,45 * m => H = 121,35 * m

    So, if I did everything correctly, and din't miss anything or made a mistake in my thinking process, question (a) is done. What I need help with is question (b). Does anyone have any ideas what the second result could possibly be?

    Thank you for your help and your time!



     
  2. jcsd
  3. Nov 9, 2016 #2

    PeroK

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    What if there were no cliff, just a boulder falling to the ground?
     
  4. Nov 10, 2016 #3
    im sorry i believe your answer is incorrect. taking your value for H, it takes it about 4 seconds to get to H/3 which will mean a speed of about 40m/s at h/3 meaning it would end at a negative altitude.

    you should start from the bottom and work your way up, i mean start where you have more data (last third of the way and work from there).
    also you shouldnt use two separate symbols for the same variable (H/3 and h, just stick with H/3)
    also dont put units in the calculus its way harder to read, specially online.
    lastly only substitute letters for values when you are going to get an actual number as a aswer till then leave it as a letter ( for example "a")

    i gave up at trying to find your mistake because of the units
     
    Last edited: Nov 10, 2016
  5. Nov 10, 2016 #4

    BvU

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    But your relevant equation (2) says U = U0 + a * t

    [edit]However, I wholeheartedly agree with WMan's advice re using symbols intead of numbers and units
     
    Last edited: Nov 10, 2016
  6. Nov 11, 2016 #5
    It'd still be the same set-up, the cliff is there just as a "background".

    Yeah, in trying to set everything apart and make it easier to understand I did end up making it needlessly complicated.

    I used 2H/3 to signify that the boulder falls from the top of the cliff (H) and arrives at h = H/3, so, in essense it traverses the two thirds of the cliff. Bearing that in mind, since my starting point is the very moment the boulder falls from the edge of the cliff, U0 is 0. So, U = 0 + (-4,9 * m/s^2) * t(2H/3)

    Anyway, yeah, I'll try it again your way. Thanks for all the tips everyone!
     
  7. Nov 11, 2016 #6

    BvU

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    Don't understand the -4.9. ##g=-9.8## m/s2
     
  8. Nov 11, 2016 #7
    Oooooh... Yeah, I get it. I got carried away by all the 1/2 * a that I messed that up.
     
  9. Nov 11, 2016 #8

    PeroK

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    If there were no cliff, and you ran time backwards, what would happen to boulder?
     
  10. Nov 11, 2016 #9
    Oh, you mean that it could keep on going "up" in a way (if we try to see it in reverse), thus falling from another height, different from the cliff's. It could be a possible explaination. Either way, as BvU pointed out, I made a mistake in calculating the velocity, so my results are off, hence the negative one. I'll run it again and I'll see what I'll get.
     
  11. Nov 11, 2016 #10

    PeroK

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    The boulder couldn't keep going up. It would reach the height of the cliff with ##0## velocity. Then, if the cliff was there, it could rest on the top of the cliff. Otherwise, where would it have come from?
     
  12. Nov 11, 2016 #11

    BvU

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    All good and well, but going via this intermediate velocity isn't making things easy for yourself. Boulder falls 3h you can give an expression for t. Boulder falls 2h you can do same. Difference is 1.5 s. You don't even get a quadratic equation. (which is what PerOK is helping you with, I think.)
     
  13. Nov 11, 2016 #12
    So you're saying I should go in reverse ? Start from the bottom and work my way up? I guess I could try it that way as well.

    PS: Some things might be getting lost in translation as english isn't my first language.

    Well, question (b) references the quadratic equation and it needs the two different results, so while skipping that part would make (a) simpler, it'd be making (b) useless.
     
  14. Nov 11, 2016 #13

    BvU

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    Meaning you could skip it :smile:
     
  15. Nov 11, 2016 #14
    Well, that's surely one way to look at it. :wink:
     
  16. Nov 11, 2016 #15

    PeroK

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    Let me give you the answer, as it's not the best problem to learn this.

    If you have an object moving under gravity, you will always get two solutions for the time when the height is ##0##. One solution (positive) is when the object will hit the ground. And one solution (negative) is when the object left the ground - if it was thrown upwards.

    In a problem such as this, the mathematics and the equations are the same, even though the object started from a point above the ground. For example, if an object starts from rest at a height ##h##, then that is the same equation as an object that was thrown up and reached height ##h## as its highest point.

    The negative solution for ##t## doesn't mean the object must have started from the ground. What it means is that if the object had been thrown upwards, that's the time when it would have left the ground.
     
  17. Nov 11, 2016 #16
    Oh, so in one case it's doing a free fall, and in the other it's the classic "I throw a tennis ball upwards" problem. Yeah, I get it now.

    Thanks a ton for the clarification, I really appreciate it. Between the translations and the fact that I'm a bit behind schedule due to the massive amount of the syllabus, I need all the help I can get.
     
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