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Basic physics logic help - kinematic equations and forces

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A small object, of mass 0.02Kg is at the top of a tall structure 160m high, it is dropped from rest.

    Ignoring air resistance..

    1. Find the speed of the object as it hits the ground
    2. find the time taken



    2. Relevant equations
    So far we have learned the equations of kinematic motion

    v=u+at
    s=1/2(u+v)t
    2s=(2u+at)t
    v^2=u^2+2as

    and forces such as

    F=ma
    Ns=mv-mu



    3. The attempt at a solution

    I get and can do the math(I think) on this particular question I am a bit puzzled as to some of the phrases and what the variables would be and then what equations and resolving thereof to use.

    The obvious ones are "s=160" "u=0" and "m=0.02". Im assuming that part 1 is asking for the final velocity (v) but cant figure out where to start.

    If the acceleration was known then I could work out what the final velocity was, however how can I find the acceleration from just that?

    Part 2 would be easy if I knew what either the acceleration or final velocity are.

    Any help appreciated....I'm sure I'm just reading it wrong or something stupid like making it more complicated than it needs to be.
     
  2. jcsd
  3. Oct 14, 2012 #2

    lewando

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    Gold Member

    If the tall structure is on planet Earth, then a = 9.81m/s2 (pointing downward). You do not need to determine this, it is a "given".
     
  4. Oct 14, 2012 #3

    Ibix

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    Science Advisor

    What did Galileo discover about things free-falling under gravity?
     
  5. Oct 14, 2012 #4
    Thanks Lewando + ibix

    I cant believe how I stupid I was, of course it is. Please forgive my ignorance I have not had any education outside of british GCSE system which was 7 years ago and have just been thrusted into a A-Level equivalent physics class and its going at a very fast pace.

    Our tutur has given g to be 9.8 for this class, so on that basis... "u" could be found as below(if i am correct)

    v^2=u^2+2as
    v^2=0+(2x9.8x160)
    v^2=3136
    √3136=56

    so part 1 is 56m/s ?
     
    Last edited: Oct 14, 2012
  6. Oct 14, 2012 #5

    lewando

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    Gold Member

    Repair your units and you will be correct.
     
  7. Oct 14, 2012 #6
    Part 2 would be the following then (if I am correct)

    2s=(2u+at)t
    320=(0+9.8xt)t
    t^2=320/9.8
    t^2=32.65
    √32.65=5.71

    t=5.71s?
     
  8. Oct 14, 2012 #7
    Thanks, I think I am getting the hang of it.

    The math/algebra is not as hard as the logic bits for me. Im a little dyslexic so sometimes I can read and re-read the question time and time again and not get what its actually asking of me.
     
  9. Oct 14, 2012 #8

    lewando

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    Gold Member

    Yes. Or since you already determined v, you could use v=u+at. With u being 0, its a faster way.
     
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