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Solving for height and gravity

  1. Jan 28, 2015 #1
    1. The problem statement, all variables and given/known data
    An astronaut on PLANET OSKI is standing on a cliff of unknown height H. The astronaut would like to ascertain the gravitational acceleration g0 of PLANET OSKI. In order to determine H and g0, the astronaut performs two experiments:

    1) FIrst she leans over the edge of the cliff and drops a rock. She notes that it takes approximately T1 = 2.0 s for the rock to land on the surface below.

    2) Next, she leans over the edge of the cliff and tosses a second rock straight upwards. The rock rises for approximately h = 1.0 m, then falls to the ground below. A time T2 = 2.34 s elapses between the moment the rock leaves her hand and the moment the rock strikes the ground.

    Find H and g0. (Assume that air resistance is negligible.)

    2. Relevant equations
    1) vf2 = v02 + 2aΔx
    2) vf = v0 + at
    3) xf = x0 + v0t + ½(a)t2
    4) Quadratic Formula

    3. The attempt at a solution
    I tried solving this three times. On the first and third tries, I failed to eliminate any variables. On the second try, I got g = 1.309 m/s and H = 2g, but this does not satisfy the astronaut's second experiment of being only 2.34 seconds long. Algebra problems with many variables like this are very time consuming and I don't have a good strategy for solving them. I spent two hours trying to tackle this problem! For this question, I would simply plug in as many known values as I am given into the equations. Then, I would start manipulating them while praying that I would somehow end up with an equation with only one variable. Is there a good strategy for this? All suggestions are welcome.
     
  2. jcsd
  3. Jan 28, 2015 #2

    Nathanael

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    Any time you blindly plug numbers into a formula, you are setting yourself up for mistakes. What if you had no answer sheet (like the astronaut)? You should still be able to come up with an answer that you have confidence in. In my opinion, the "kinematics equations" are entirely useless and can cause more trouble than they're worth. The only kinematics equations you need to know are [itex]v=\frac{dx}{dt}[/itex] and [itex]a=\frac{dv}{dt}=\frac{d^2x}{dt^2}[/itex]

    All of the information in the problem is summarized in this picture:
    trivial.png
    All you need to know is that the slope of the graph is g0. (And how to find the area of a triangle.)

    Now write a few equations describing the area of each triangle.
     
  4. Jan 28, 2015 #3
    Thank you for your suggestion, here is what I got:
    Area of a triangle is (1/2)LW, so when she drops the rock it's H = 0.5 * 2 * Vf.
    When she throws the rock it is H+1 = 0.5 * (Time need to fall H + 1 meters) * Vf. This final velocity is different from the case where the rock falls.
    I also know that the line for the two graphs is y = mx + b. So it's y = -gt (where t = x axis) for the falling rock and y = -gt + V0.
    Where should I go from here?
     
  5. Jan 29, 2015 #4

    Nathanael

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    Now write Vf in terms of g. This gives you one constraint on H and g (the other constraint comes from the other graph).

    Right, so now write this final velocity in terms of g and V0

    Also, break up the 2.34 second interval into two times, T1 and T2, such that T1 is the time from when the rock is released to when it has zero velocity. Then T2 will be the time from when it has zero velocity (at it's max height) to the time when it hits the ground. Then write an equation for the remaining triangle (the one with area 1 meter).
     
  6. Jan 29, 2015 #5
    I get H = 0.5 * 2 * gt --> H = 2g
    For the thrown rock, I get Vf = V0 + gt --> Vf = g(2.34). This is the interval from max height to ground.
    The other triangle has area = 1 = 0.5(Time to reach max height of 1 meter)V0
     
    Last edited: Jan 29, 2015
  7. Jan 29, 2015 #6
    How do I get rid of an attachment?
     
  8. Jan 29, 2015 #7

    Nathanael

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    I'm not sure why you dropped the V0, it's not zero. (Perhaps just a typo?) It should be Vf=V0-2.34g (g is negative because the slope is negative)

    This comes from the definition of slope, [itex]\frac{Δy}{Δx}=-g[/itex] therefore [itex]Δy=-gΔx[/itex]
    (g is only negative because it's sloping downwards, that is, a positive change in x gives a negative change in y)

    In this case, y is velocity and x is time, so slope is actually [itex]\frac{Δv}{Δt}[/itex] (which of course has units of acceleration)

    Right. I prefer to call it T1 and T2 (it's easier to read). Do you know any relationship between T1 and T2? That's the final key and then it's just algebra. (You can then eliminate V0 from the two equations for the 1 meter triangle and the (H+1) meter triangle.)

    Click the edit in the bottom left corner of your post; you should just be able to delete it that way.
     
    Last edited: Jan 29, 2015
  9. Jan 29, 2015 #8

    Nathanael

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    P.S.
    Doing it this way is equivalent to using the "kinematics equations." I just wanted you to do it this way because you said you felt like you were blindly manipulating equations.
    (Plus, I never memorized any "kinematics equations" so I prefer to solve the problem visually. You can derive every single "kinematics equation" from triangles like this, so if you understand the visual approach, then there's no need to memorize anything!)
     
  10. Jan 29, 2015 #9
    Why isn't V0 zero? In your last post, you wrote that Vf=V0-2.34g. If Vf = the velocity at 2.34 seconds, shouldn't the initial velocity be zero because the rock is at its maximum height. On the other hand, if we take Vf to be the velocity at the maximum height, then V0 = nonzero and Vf = zero. Correct me if I'm wrong but it seems that it's not possible to have both Vf and V0 in the same equation....
     
  11. Jan 29, 2015 #10

    Nathanael

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    You could use your equation where V0 is zero (at the maximum height) but then you would not want to use the full 2.34 seconds! You would just want to use T2 (so it would be Vf=T2g)
    (I hope this explains the confusion.)

    I'm thinking it might be a little algebraically easier later if you use the full 2.34 seconds and the initial velocity.
    (But I haven't gone through any of the algebra myself so I could be dead wrong about that.)
     
  12. Jan 29, 2015 #11
    You can start the problem this way :

    ##\left\{
    \begin{array}{}
    a(t) = -g_0 \\
    v(t) = v(0) - g_0t \\
    y(t) = y(0) + v(0)t - g_0 \frac{t^2}{2}
    \end{array}
    \right.
    ##

    1) First part tells you: ##y_1(T_1) = 0,\ y_1(0) = H,\ v_1(0) = 0 ##
    2) Second part tells you:
    2.1) ##y_2(T_2) = 0,\ y_2(0) = H##.
    2.2) There is a time ##T_3## for which the rock is as the top of its trajectory which means that ##v_2(T_3) = 0## and ##y_2(T_3) = H+h##
     
    Last edited: Jan 29, 2015
  13. Feb 12, 2015 #12
    http://photo1.ask.fm/927/001/697/-359996973-1shmnb5-bn87st5cd32olk/original/IMG_40971Copy.jpg [Broken]
    Sorry it's tilted but this is my work, but it is not correct.
     
    Last edited by a moderator: May 7, 2017
  14. Feb 12, 2015 #13

    haruspex

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    It's also hard to read, partly because the logical flow is not clear, partly because you plug in numbers instead of keeping everything symbolic.
    There are five SUVAT equations, each involving four of the same five variables: initial and final speed, distance, time, acceleration. In this problem, the only interesting speeds are the ones known to be zero, so pick the equation which only involves one speed variable and use it for each of the three motions: drop from cliff, toss up from cliff to highest point, fall from highest point. There will be three unknowns: height of cliff, acceleration, time to reach highest point. Three equations, three unknowns.
     
  15. Feb 15, 2015 #14
    This is my work
    In the first scenario, the rock simply falls.
    Taking down as positive, we get:
    ## H = -2g ##

    In the second scenario, the rock is thrown
    Looking only at the first part of the motion (end point is maximum height).
    ## v_0 = \sqrt {2g} = gt_1 ##
    Where t1 + t2 = 2.34 seconds.

    Looking at the second part of the motion (starting point is max height)
    ## v_f = \sqrt {-2gH - 2g} = gt_2 ##

    Combining all of the information:
    ## v_0 + v_f = gt_1 + gt_2 ##
    ## = v_0 + v_f = 2.34g ##
    ## = 2.34g = \sqrt {2g} + \sqrt {-2gH - 2g} ##
    Note that ## H = -2g ##
    g = wrong answer of 14.178 m/s2
     
  16. Feb 15, 2015 #15

    haruspex

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    Do v0 and vf have the same sign?
    PLEASE keep everything symbolic: use h, T1, T2 as given until the final equation is reached.
     
  17. Feb 15, 2015 #16
    v0 and vf have opposite signs. v0 is negative while vf is positive.
    First half of motion
    ## v_f = v_0 + at ##
    ## 0 = -v_0 + gt_1 ##
    ## v_f^2 = v_0^2 + 2 a \Delta x ##
    ## 0 = -v_0^2 + 2g ##

    2nd half
    ## v_f = v_0 + at ##
    ## v_f = 0 + gt_2 ##
    ## v_f^2 = v_0^2 + 2 a \Delta x ##
    ## v_f^2 = 2g(-H-1) ##
     
  18. Feb 16, 2015 #17

    haruspex

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    As I mentioned in post #13, you don't care about the speed with which the rock was thrown up or the speed with which it hits the ground, so you're making life hard by selecting the equations that involve initial and final speeds. Better to use ##s=v_0t+\frac 12 at^2## and ##s=v_ft-\frac 12 at^2##. In each case, you know one speed to be zero, so you have three occurrences of ##s=+/-\frac 12 at^2##.
     
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