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What acceleration did she experience during the collision?

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A very silly person, intent on catching pigeons on the roof of an apartment building, trips and falls a distance of 44.6 m. She lands on a metal garbage can, crushing it to a depth of 0.453 m and walks away without having been hurt. What acceleration did she experience during the collision?

    And also...
    A parachutist jumps from an airplane and freely falls y=49.1 m before opening his parachute. Thereafter, he decelerates at a=1.92 m/s2. As he reaches the ground, his speed is 2.90 m/s. How long was the parachutist in the air?

    My answer was 11.73 s, which was correct.

    The second part of the question is: At what height did the parachutist jump from his plane? This is the one I can't figure out.


    2. Relevant equations
    s_f = s_i +v_it + 0.5at^2


    3. The attempt at a solution
    I couldn't get anywhere with them!
     
  2. jcsd
  3. Sep 16, 2007 #2

    learningphysics

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    1) What do you need for the acceleration?

    2) you know the first part is 49.1m, you need the second part of the fall. the sum gives you the height. Any ideas? what equations can you use?
     
  4. Sep 16, 2007 #3
    I need the time for the acceleration... I tried and I got 3.02s, but that didn't work out in my equations that followed.

    The second part of the fall... wouldn't be 0.453m? I tried using the equation I stated above with -9.8m/s^2, but that didn't work. I also tried adding the two heights together and had no luck either.
     
  5. Sep 16, 2007 #4

    learningphysics

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    can you show all your calculations... then we can point out where you're going wrong...
     
  6. Sep 16, 2007 #5
    FOR THE PARACHUTIST QUESTION:

    I got 31.02m/s as Vf for the first segment:

    Vf^2 = Vi^2 +2ad
    where Vi = 0 m/s, a = 9.8 m/s^2, and d = 49.1 m

    but when I tried using that for the second part, I got the wrong answer. I got an answer of 248.39 m from:

    displacement = (2.90^2 - 31.02^2) / (2 x 1.92)


    FOR THE FALLING WOMAN QUESTION:

    Vf^2 = Vi^2 + 2ad
    where Vi = 0 m/s, a = 9.8 m/s^2, and d = 44.6 m

    And then:

    a = (Vf^2 - Vi^2) / (2 x d)
    where Vf = 0 m/s, Vi = 29.57 m/s, and d = 0.453 m.
     
  7. Sep 16, 2007 #6

    learningphysics

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    So the height is 49.1 + 248.39 = 297.49m ... looks right to me the computer didn't accept 297.49?

    One thing I'd be careful off... be consistent with your signs... it seems like you're taking downwards as positive... and using d = 0 at the top with increasing d going downwards... that's fine, but with that convention, you should use a = -1.92 (seems like you just ignored the minus sign).

    This also looks right... I'm getting a = -964.86m/s^2. Is that what you submitted? Maybe you need to take up as positive, down as negative, so acceleration is 964.86m/s^2 ?
     
  8. Sep 16, 2007 #7
    AHH THANK YOU!

    For the first one, I kept forgetting to add the two displacements together.

    And for the second one, it should have been positive!

    THANKS AGAIN!
     
  9. Sep 16, 2007 #8
    Actually, hold on. I'm getting 965.10 m/s^2.
     
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