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A climber throws 2 stones 1.00s apart and they hit at the same time

  1. Sep 25, 2007 #1

    ~christina~

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    A climber throws 2 stones 1.00s apart and they hit at same time h=50ft

    height of cliff = 50ft I think there is enough info now...

    1. The problem statement, all variables and given/known data
    A climber climbs a cliff overhanging a calm pool of water and throws 2 stones vertically down 1.00 s apart. The climber then observes that they cause a single splash. The first stone has a initial v of 2.00m/s

    a) how long will it take after the release of the stones do the 2 stones hit the water.

    b) what is the inital velcocity of 2nd stone if they are to hit simultaneously?

    c) What is the speed of each at the instant the 2 stones hit the water?


    2. Relevant equations

    If I'm not incorrect in which eqzn I used...

    then Position as a function of velocity and time...

    xf= xi + 1/2(vxi + vxf)t



    3. The attempt at a solution

    a.)
    d= 50 m ===>since it is starting from 0 if I'm not incorrect and flung downward then the final distance is 50 so it's negative 50...I think

    diference in time thrown is = 1.00s

    vi stone 1 = 2.00m/s

    a= -9.80m/s^2


    ___________________
    xf= xi + vxi (t) + 1/2 (ax)(t^2)

    -50m= 2.00t + .5(-9.80m/s^2)

    0= 50m+ 2.00t -4.9t^2

    through quadradic formula I found that

    t= -2.00+/-rad (2.00)^2 - 4*(-4.9)*50
    -9.8

    and for that I found that t= 3.40s since the neg can't be used as a valid time
    _____________________

    Then for b)

    I equated the 2 equations leaving out vxi for one side and for the time since they say a difference of 1.00s I went and added it to the time found for the first stone and used that but I'm not sure about that. (Assuming that the sencond stone was thrown after the first one) Is acceleration -9.8? since the person is throwing the stone not dropping it I wasn't sure about that.

    xi-xf + vxi*(t)+ 1/2*ax*t^2 = xi-xf + vxi*(t)+ 1/2*ax*t^2
    stone 1-------------------------stone 2

    -50 + 2.00m/s*3.40s + 1/2*-9.80m/s^2*(3.40s)^2= -50 + vxi*4.40s+ 1/2*(-9.80m/s^2)*(4.40s)^2

    -99.84= -144.864 + vxi*4.40s
    45.024 = vxi*4.40s
    vxi= 10m/s


    _____________________________

    For C.) I don't know how to approach that..which equation do I use?
    I know it's a kinematic one but I'm not sure which to use....





    Thanks :confused:
     
    Last edited: Sep 25, 2007
  2. jcsd
  3. Sep 25, 2007 #2
    Not enough information is given. Enough is given to figure out the velocity and position of the first stone the instant the second stone is thrown (I do that below), and after that point in time you can ignore gravity to find when they meet because they are free-falling (in their own inertial reference frame). But you still can't figure out when they meet because you don't know their relative velocity. Picture the extremes when the second stone is thrown:

    1. The second stone is fired downward at an immense velocity, causing the stones to meet almost instantaneously. In this case, the time in the air is 1 second.

    2. The second stone is thrown downward with a velocity just barely faster than that of the first stone, and it takes ages for them to meet.

    At least you can figure out the position and velocity of the first stone when the second is thrown:

    First, I take it you mean the first stone has an initial v of -2m/s (negative because its going down). So...

    Vfinal = Vinitial + acceleration * time
    V1 = -2m/s + -9.81m/s * 1 = -11.81m/s

    And the position (relative to the climber) of the first stone when the second stone is thrown is...
    Yfinal = 1/2 acceleration * time2 + Vinitial * time
    Y1 = 1/2(-9.81) * 12 + -2 * 1 = -6.905

    So when the second stone is thrown, the first stone is falling at 11.81m/s and is 6.9 meters below the climber. At the point the second stone is thrown, the two stones are free falling and we only need to know when they meet, so we don't need to worry about gravitational acceleration. However, when they meet depends on the relative velocity of the first stone, so we don't know when they meet, and thus don't know when they hit, and thus don't know the answers to any of the other questions.
     
  4. Sep 25, 2007 #3

    andrevdh

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    They share the area a and b. So area c and d need to be the same as area e (use the known value of the gradient). Hope it helps!
     

    Attached Files:

  5. Sep 25, 2007 #4

    andrevdh

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    Another approach that generates an additional relationship (in which case you have two equations with two unknowns):

    You can calculate the initial distance, [tex]d_o[/tex], between the two stones and the speed of the first one at the stage that the second one is thrown. Also the relative speed of the two stones stays the same throughout the motion, that is if their initial speed are [tex]u_1,\ u_2[/tex], it follows from

    [tex]v_1 = u_1 +gt[/tex]

    and

    [tex]v_2 = u_2 +gt[/tex]

    that

    [tex]v_{rel} = u_2 - u_1[/tex]

    which is not time dependent. Therefore we can say that

    [tex]v_{rel} = u_2 - u_1 = \frac{d_o}{T_{rel}}[/tex]

    where [tex]T_{rel}[/tex] is the elapsed time since the second stone were cast until they met (at the surface of the pool).
     
    Last edited: Sep 25, 2007
  6. Sep 25, 2007 #5

    ~christina~

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    I don't get it so you say that I can't find out the time they hit the water....

    for the relative velocity ..... how would I find the value for d and vrel if they are unknowns??

    And why can't I use the equationI used before???
    I have a difficult time choosing equations.
     
  7. Sep 25, 2007 #6

    ~christina~

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    I forgot to mention that the height of the cliff is 50ft
     
  8. Sep 26, 2007 #7

    andrevdh

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    I was trying to solve the problem without the height. When the height is given the problem becomes much easier!

    In post #1 you say:

    xf= xi + vxi (t) + 1/2 (ax)(t^2)

    -50m= 2.00t + .5(-9.80m/s^2)

    the intial speed is also downwards so it should also be negative (you forgot the t squared in the last term). If you choose downwards as positive all three terms are positive (you also said that the distance is in feet?). This will enable you to solve the quadratic for the time, say T, it takes the first stone to hit the water.

    for part b the second stone takes one second less to hit the water, T - 1 (it was thrown one second after the first one - see the attachment).

    They both travel the same distance in these times. So you can equate the two

    ut + 1/2 a t^2

    for both stones. For the second stone u is unknown, and it is the only unknown in the resulting equation.
     

    Attached Files:

    Last edited: Sep 26, 2007
  9. Sep 26, 2007 #8

    ~christina~

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    xf= xi + vxi (t) + 1/2 (ax)(t^2)

    -50m= -2.00t + .5(-9.80m/s^2)t^2

    t= 3.00s

    so since t-1 is the 2nd stone t-1= 3.00s-1= 2.00s

    and to find vxi I plugged in for the unknown vxi


    0+ vxi*2+ 1/2(-9.80)(2^2)= 0 + (-2* 3) + 1/2 ( -9.8)(3^2)
    vxi= -15.25m/s

    Is this alright? It looks kind of funny (the number for the velocity I got for the 2nd stone)


    _______________________________________________
    For C.) where I have to find the speed of each stone at the instant the 2 hit the water?

    I think that I plug the numbers I got into the eqzn vxf= vxi + ax*t

    If I'm correct then It would be

    vxf= vxi + axt

    stone 1: t= 3

    (-2) + (-9.80* 3)= -31.4m/s

    stone 2: t= t-1 = 3-1= 2s

    (-15.25) + (-9.8)(3-1)= -34.85m/s

    ~Thank You~
     
  10. Sep 27, 2007 #9

    andrevdh

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    I also get 3.00 seconds for the time for the first stone to reach the surface of the pool.

    0+ vxi*2+ 1/2(-9.80)(2^2)= 0 + (-2* 3) + 1/2 ( -9.8)(3^2)
    vxi= -15.25m/s

    seems ok, but you could have made it a bit easier on yourself by just:

    -50 = 0 + vxi*2+ 1/2(-9.80)(2^2)

    then I get -15.2 m/s on the dot. One would expect that it would need quite a large speed to catch up with the first stone in only two seconds.

    The intial speeds seems fine.

    On further investigation the relative speeds are not the same at the beginning and end

    beginning: 15.2 - 2 = 13.2 m/s
    end: 34.8 - 31.4 = 3.43 m/s

    It seems it is only true that the relative speed of the objects will be constant throughtout the motion if they are launched simultaneously. That is my "proof" in #4 should include the time difference:

    [tex]v_1 = u_1 + gt[/tex]

    [tex]v_2 = u_2 + g(t + \delta)[/tex]

    so that the relative speed is time dependent:

    [tex]v_{rel} = u_2 - u_1 + \delta t[/tex]
     
  11. Sep 27, 2007 #10

    ~christina~

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    So the last part I did was incorrect?

    I included the t-1 in the equation though...
     
  12. Sep 27, 2007 #11

    andrevdh

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    No, it is correct. My previous remark that the relative speed of the two stones should stay the same was wrong. That is what I was trying to explain.
     
  13. Sep 29, 2007 #12

    ~christina~

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    Oh okay...Thanks andrevdh! :smile:
     
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