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Free Fall physics homework

  1. Nov 19, 2005 #1
    Pepe, the clown, is jumping on a trampoline as Babette, the tightrope walker, above him suddenly loses her balance and falls off the tightrope straight toward Pepe. Pepe has just started upward at 15.0 m/s when Babette begins to fall. Pepe catches her in midair after 10.0s. a) How far has Babette fallen when she is caught by Pepe? b)What is Babette's velocity at the time of contact? c) What is Pepe's velocity at the time of contact? d) How far above the trampoline was Babette before she fell?
    a) d= -491m
    b) I used v final^2=v initial^2 + 2ad, but the answer key uses v final=v initial +ad. Why doesn't v final and v initial have squares? The answer comes out to be the same as mine except mine is missing the negative, so i'm supposing the ans key is right, but i don't understand how vf^2 andvi^2 can change into vf and vi without the other parts of the equations being square rooted. :confused: :confused:
     
    Last edited: Nov 19, 2005
  2. jcsd
  3. Nov 19, 2005 #2

    Fermat

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    I haven't checked you answers, but v final=v initial +ad is wrong. It should be v final=v initial +at, where t is the time duration of the fall.
    If you check the units of the bad eqn, ad has units of m²/s² although vi and vf both have units of m/s
     
  4. Nov 19, 2005 #3

    Fermat

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    Here is the derivation of your eqns of motion.

    Let u be the initial velocity, v be the final velocity, a be the constant acceleration and s be the distance travelled.
    If a body starts at an initial speed of u and has a constant acceleration of a, then its speed increases by a for every second that passes. This gives you a basic eqn of motion,

    v = u + at
    ========
    or
    (v-u) = at

    If a body has an initial speed of u and a final speed of v, then its average speed is given by,

    vav = (v+u)/2

    distance travelled is then,

    s = vav.t
    s = (1/2)(v+u).t
    =============
    or
    (v+u) = 2s/t

    From the two re-arranged eqns,we get

    (v-u)(v+u) = at.2s/t
    v² - u² = 2as
    or
    v² = u² + 2as
    ===========

    vav = (v+u)/2
    vav = (u + at+u)/2
    vav = u + (1/2)at

    but

    s = vav.t
    s = (u + (1/2)at).t
    s = ut + (1/2)at²
    ============
     
  5. Nov 19, 2005 #4
    ok, so I see now, it is +at, not ad, so it makes sense now:
    b) vf=-98.1m/s

    and for c), i think this should be correct:
    vf=at+vi
    vf=(-9.81m/s^2)(10.0s)+15.0m/s
    vf=-83.1m/s

    but for d), I'm having a bit of trouble:
    the formula d=vit+(1/2)at^2 should be used, i expected that Babette's vi was 0 because she fell, but for the ans key, it says vi is 15.0m/s, isn't that supposed to be Pepe's vi though? would it just be an error in the ans key, or am i missing smthg?
     
  6. Nov 19, 2005 #5

    daniel_i_l

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    For d) you want to know how far the clown travelled up (from the trampoline) , not how much the tightrope walker fell. Also, don't forget the (-) in the equation (travelling up).
     
  7. Nov 19, 2005 #6
    for d), isn't it asking how far above the trampoline was Babette before she fell, and Babette is the one falling, so what does this question have to do with Pepe jumoing on the trampoline?
     
  8. Nov 19, 2005 #7

    Fermat

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    You know how far babette fell from the tight-rope, when she was caught by pepe. Now you have to find how high pepe got to when he caught babette, then add the two distances together to find out how high the tight-rope is.
     
  9. Nov 19, 2005 #8
    thanks, I thinki got it now
     
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