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Mechanics 1 (M1) - Vertical motion under gravity

  1. Sep 26, 2008 #1
    1. The problem statement, all variables and given/known data
    A girl throws a ball vertically upwards with speed 8ms^-1 from a window which is 6m above horizontal ground.
    (a) Find the greatest height above the ground reached by the ball. (COMPLETE)
    (b) Find the distance below the window of the point where the balls meet.

    [​IMG]

    2. Relevant equations
    Vertical motion under gravity (ignoring air resistance)- Releasing objects
    v=gt
    v^2=2gh
    h=1/2gt^2

    Vertical motion under gravity (ignoring air resistance)- Projecting objects
    v=u-gt
    v^2=u^2-2gh
    h=ut-1/2gt^2

    Where:
    v=final velocity
    u=initial velocity
    t=time
    g=acceleration due to gravity (count as 9.8ms^-2)
    h=height from ground(a.k.a. distance)

    3. The attempt at a solution
    Well I have done the first part...
    (a) 2gh=u^2-v^2
    h=(u^2-v^2)/2g
    =(8^2-0)/2*9.8
    =3.27m
    6+3.27=9.27m (FINAL ANSWER)
    However for (b) I do not know where to start. I know the answer is 0.104m (checked the answers) but I have no idea how to arrive there, I have simply done scribbles all over the place:
    Attempt 1
    [​IMG]

    Attempt 2 (Using previous info)
    [​IMG]

    I hope someone can tell me where I am going wrong or give me a few tips on what to do and I should be able to calculate the rest.

    P.S. Thanks in advance! I'm not sure if this is the right section as this unit is part of the maths AS course but mechanics is mainly physics so I placed the question here - my apologies if it is in the wrong section.
     
    Last edited: Sep 26, 2008
  2. jcsd
  3. Sep 26, 2008 #2

    LowlyPion

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    Homework Helper

    Re: Mechanics 1 (M1)

    Welcome to PF.

    The first part looks good.

    The second part requires that you develop 2 equations, one for the motion of each ball.
    You know that they will meet at the same time.
     
  4. Sep 26, 2008 #3
    Thanks for the welcome!

    I really can't find the answer though lol.

    I will rest my head for tonight and try this problem again tomorrow, thanks for now.
     
  5. Sep 26, 2008 #4

    LowlyPion

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    Homework Helper

    While you are resting think about where the first ball will be and how fast it is going when the second ball is dropped.

    Get a good sleep and maybe it will occur to you in a dream.
     
  6. Mar 22, 2010 #5
    Hi, I think I might have the solution for your problem.
    First of all, I think that the answer s = 0.104m is not really that accurate.. I tried plugging in your answer and it wont meet at that distance below the window..

    Anyway, here is my solution, I used g = 9.81ms^-2

    First ball,
    s = ut + 1/2 at^2
    s = 8t - 4.905t^2

    For second ball,
    Using the same equation,

    s = 0 + 4.905( t -1.5 )^2
    This is because since the second ball is always ahead of the first ball by 1.5 s.

    Now, this equating both equation, and putting in mind that the we let the motion of the first ball that is thrown upwards to be negative and the motion of the second ball downwards to be positive displacement.

    s (1) = s (2)

    -(8t - 4.905t^2) = 4.905 ( t-1.5 )^2
    Expanding the equation,

    You will get,
    6.715t = 11.036
    Hence, t = 1.6435s

    Now plugging back to either equation,
    Lets say the first equation,

    s = -8(1.6435) + 4.905(1.6435)^2
    s = 0.101 m

    And you will get the same if you put it in equation 2..
    So both equation are satisfied..
    Even so, if i used g =10 I still will not get 0.104m

    Anyway i hope that you are happy with my solution and I hope you understand this..
    All the best,
    By Victorian91..
    Physics rocks..
     
  7. Nov 4, 2010 #6
    This question too has been bothering me but I've just managed to sort it out ! Here's my solution.

    Stone A:
    s=s1m
    u=-8ms-1
    v=?
    a=9.8ms-2
    t=t

    Stone B:
    s=s2m
    u=0ms-1
    v=?
    a=9.8ms-2
    t=t-1.5


    So from using equations of motion on Stone A:
    s=1/2(u+v)t2
    =>s1=-8t+1/2(9.8)t2
    1=>s1=-8t+4.9t2

    And using the eqations of motion on Stone B:
    s=1/2(u+v)t2
    =>s2=0t+1/2(9.8)(t-1.5)2
    =>s2=4.9(t-1.5)2
    =>s2=4.9(t2-3t+2.25)
    2=>s2=4.9t2-14.7t+11.025


    Make the equations equal to each other in order to find the time in which the balls meet:
    1=2
    =>-8t+4.9t2=4.9t2-14.7t+11.025
    =>-8t=-14.7t+11.025
    =>6.7t=11.025
    =>t=11.025/6.7
    =>t=1.645522388
    =>t=1.65s


    Now substitue t=1.65s into equation, 1:
    =>s=-8t+4.9t2
    =>s=-8(1.65)+4.9(1.65)2
    =>s=0.104m

    Although I've managed to do it correctly on paper, I believe I may have made many sign errors on here so watch out !! If you spot what they are feel free to tell me so I can correct it. (:

    Also Victoria91, you made a mistake in your calculation of t, it is slightly off, which is why your answer for s is also slightly off.
     
    Last edited: Nov 4, 2010
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