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A daring ranch hand.

  1. Jul 31, 2007 #1
    A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 10.0 m/s and the distance from the limb to the level of the saddle is 3.00 m. (a) What must be the horizontal distance between the saddle and the limb when the ranch hand makes his move? (b) How long is he in the air?

    I have the basic equations of mechanics:

    v = u + at
    s = ut + .5a(t^2)
    v^2 = u^2 + 2as

    However, I took a different approach to the problem. I labeled the level of the limb with s = 3.00, v = 0, and a = -9.8. The level of the horse would be s = 0, v = ?, and a = -9.8, in terms of the rancher. Using integration and beginning with a = -9.8:

    v(t) = -9.8t + c

    velocity equals zero when time equals zero, so:

    v(0) = -9.8(0) + c = 0
    c = 0

    Thus, equation for velocity is: -9.8t

    Another integration to find the equation for position was used:

    s(t) = -4.9t^2 + c

    The position of the rancher is 3 when time is zero:

    s(0) = -4.9(0)^2 + c = 3
    c = 3

    Equation of position becomes: -4.9t^2 + 3

    Before going any further, I wanted to make sure I was going about this problem the right way.

    If my train of thought is correct, I planned to set the position equation I acquired equal to zero to get the time it takes for the rancher to jump from the limb to the horse's level. That answers part (b) of the problem. Once I figure that out, I could use a proportion to judge how far away the horse should be from the point of being directly below the rancher, since velocity is a rate:

    10.0 m = ________________ X _________________________
    1 sec [however many seconds the rancher was in the air]

    If this method works, then great! If not, I would like it if someone could possibly correct my logic in terms of the methods I am using.

    Also, I am not exactly sure how to go about doing this problem in terms of...well, physics (and using the equations they gave me). I went about it by just remembering what I was taught in Calculus I last semester, but my physics chapter did not go about solving problems in the manner I thought of.

    Thanks in advance! :smile:
  2. jcsd
  3. Jul 31, 2007 #2


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    Your reasoning is entirely correct.

    Of course, you could have directly got the vertical equation by using the general formula

    [tex] y(t) = y_i + v_{0y} t + \frac{1}{2} a_y t^2 [/tex]
    and setting [itex] y_i =3, v_{0y} =0, a_y = -9.80 m/s^2 [/itex]
  4. Jul 31, 2007 #3
    ...that works too.

    And it's a whole lot easier.

    I guess I just saw it better with all of the integrations and proportions.

    Thank you!
  5. Jul 31, 2007 #4


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    Your reasoning was entirely correct. And the second part where you use a rule of 3 (proportions) is valid but you could as well have instead tsaid that you use, along x, [itex] x(t) = x_i + v_{xi} t [/itex] Setting x_i =0 and plugging in the time you found before gives the same answer as what you get using proportions.

    The way I showed, using the equations of kinematics, is the standard approach but your way was correct.

    You are welcome.
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