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Rectilinear Motion (Object dropped 500m, will it burst?)

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    An object is dropped from a height 500 m above the ground. This object is designed to withstand an impact velocity of 100 m/s. Will it burst?

    2. Relevant equations

    gravitational force: 9.8 m/s

    3. The attempt at a solution

    s(t) = 500 m
    a(t) = 9.8 m/s
    v(t) = ?

    a(t) = 9.8
    Therefore, v(t) = 9.8t + C

    What do I do from here? How do I find for C?
     
  2. jcsd
  3. Mar 22, 2010 #2

    rock.freak667

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    You have s=500 and a=-9.81, they tell you that it is dropped from that height (so what is the initial velocity u?)

    What equation of motion relates, v with u,a and s ?
     
  4. Mar 22, 2010 #3
    I think you are on the right path.
    v(t) = 9.8t + C
    but you know something about the velocity at t = 0! This condition will give you C.

    After you got v(t), you can get s(t) with the same method. Keep in mind the direction of v! Finally you can calculate the time for the object to hit the ground and its velocity.
     
  5. Mar 22, 2010 #4
    Thanks for your replies and guidance rock.freak667 and justsof.

    Ok, here is another attempt based on your what you said:

    Now the v = g*t

    a(t) = -9.8 m/s
    v(t) = -9.8t + C
    v(0) = -9.8(0) + C = 0
    C = 0

    v(t) = -9.8t + 0

    s(t) = (-9.8t^2)/2 + D

    (-9.8t^2)/2 + D = 500

    (-9.8(0)^2)/2 + D = 500

    D = 500

    (-9.8t^2)/2 + 500
    I am really confused at this point
     
  6. Mar 22, 2010 #5

    rock.freak667

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    Right so then you have s(t)=500-(9.8/2)t2

    s(t) is measured from the ground (at 0 displacement) to the object. So you want to find the 't' at which the object hits the ground.

    So you want s(t) to be equal to what value?
     
  7. Mar 22, 2010 #6
    I want s(t) to be equal to 500 m? At this point, I am just guessing but...
     
  8. Mar 22, 2010 #7

    rock.freak667

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    No s is measured from the ground to the object. So at the ground, shouldn't s=0?
     
  9. Mar 22, 2010 #8
    Ok, yes you are right, the ground is 0, that makes sense,

    But my question is, what rules should I follow when working out problems like this? My problem is, the only rules I know are: a(t) = v'(t) = s"(t), but I don't know when to apply them.

    Ok, since the object has hit the ground now, we have the current location, so do we have to differentiate to get the velocity like this:

    s(0) = 500 - (9.8(t)^2)/2
    therefore v(t) = 19.6 This does not make any sense, I'll try again


    or do we just plug in the 0 into the formula like this: s(0) = 500 - (9.8(0)^2)/2 = 500 This also does not make any sense to me

    s(0) = 500 m
     
  10. Mar 22, 2010 #9

    rock.freak667

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    0 = 500 - (9.8(t)^2)/2, what is the value of t?

    then just use v=u+at.


    But there was a much simpler way to do this question, where you did not have to go through finding the time.
     
  11. Mar 23, 2010 #10
    Isn't t = 0?
     
  12. Mar 23, 2010 #11

    rock.freak667

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    the 't' you will use there is time taken for the object to fall 500m
     
  13. Mar 23, 2010 #12
    Thanks for your patience rock.freak667.

    I obviously need to practice problems like this. Can you recommend any basic tutorials on line similar to this?

    This is pressure!

    Ok, Im I on the right track now:

    0 = 500 - (9.8t^2)/2

    0 = 250 - 9.8(t)^2

    -250/-9.8 = t^2

    square root of 25.5 = t

    5 = t

    Object drops at 5m/s ??
     
  14. Mar 23, 2010 #13

    rock.freak667

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    you forgot to divide by the 2, 500-(9.8/2)t2=0, t is not 5.
     
  15. Mar 23, 2010 #14
    I'll try again:

    0 = 500 - (9.8t^2) /2
    0 = 500 - 4.9t^2
    -500/4.9 = t^2
    -102 = t^2
    square root of -102 = t
    10.09 = t

    Now as you said, v=u+at
    v = ?
    U = 0 (initial velocity)
    a = 9.8 m/s
    t = 10.09 seconds

    Therefore, v = 0 + 9.8(10.09) = 98.88 m/s
    Is this correct?
     
    Last edited: Mar 23, 2010
  16. Mar 23, 2010 #15

    rock.freak667

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    and thus the velocity at t=10.09 is ?
     
  17. Mar 23, 2010 #16
    Jezzzz, did I still get it wrong!

    The question asks whether the can, which can withstand impact velocity of 100 m/s, will burst.

    So isn't the answer no it won't burst because its velocity is 98.88 m/s (see working below)

    0 = 500 - (9.8t^2) /2
    0 = 500 - 4.9t^2
    -500/4.9 = t^2
    -102 = t^2
    square root of -102 = t
    10.09 = t

    Now as you said, v=u+at (here is your formula)
    v = ?
    U = 0 (initial velocity)
    a = 9.8 m/s
    t = 10.09 seconds

    Therefore, v = 0 + 9.8(10.09) = 98.88 m/s
    Is this correct?
     
  18. Mar 23, 2010 #17

    rock.freak667

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    Yes that is correct.

    An alternative way was to just use a kinematic equation that involves v,u,g and s.

    v2=u2+2g(s-s0)
     
  19. Mar 23, 2010 #18

    Dick

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    sqrt(-102) isn't 10.09. You moved stuff around wrong in doing the algebra. Just a sloppy mistake, right?
     
  20. Mar 24, 2010 #19
    I'm not no physics expert, but, can't you just solve for vf and if it is less then 100m/s then it will not burst...i solved for vf and got 98.99m/s...this shows that the object will not burst because the impact velocity is less then 100m/s
     
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