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Another Kinematics problem

  1. May 28, 2006 #1
    Alice, Bob and Daniel are standing on the edge of a canyon which is 100m deep and 20m wide. On the opposite side of the canyon there is a 10m high cave, 15m below the top of the cliff. They each throw a rock horizontally with the following initial speed to see who can get it into the cave.

    Alice rock speed 5m/s
    Bob rock speed 10m/s
    Daniel rock speed 20m/s

    for this qn, i do not understand why initial velocity is 0....hmmm ??? becoz only with the initial speed as 0 can i work out the correct ans...if i were to use the values given, i will get a wrong ans...
     
  2. jcsd
  3. May 28, 2006 #2
    The initial velocity in the vertical direction is 0 since the stones are being thrown horizontally.
    In the horizontal direction, the stone always has the velocity as given in the question .
     
  4. May 28, 2006 #3

    Hootenanny

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    Ignoring air resistance of course :wink:

    ~H
     
  5. May 28, 2006 #4
    And various other factors, of course
     
  6. May 28, 2006 #5

    Andrew Mason

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    Work out the equation for speed in terms of the vertical drop distance over the 20 m. width. Then work out the range of permitted speeds that would result in a drop of between 5 and 15 m. (I assume the floor of the cave is 15 m. below the top).

    AM
     
  7. May 28, 2006 #6
    hmmm....but the diagram that is drawn on the paper shows that when they throw the rock, it will form an angle...since they are at the top of the cliff, the cave is opposite them and slightly below them...

    when solving this sort of qn, do i have to factor in the angle? or i can juz treat the qn independently in terms of x and y axis
     
  8. May 28, 2006 #7

    Hootenanny

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    You would need factor in the angles when resolving the horiztonal and vertical components. For example, if a ball is thrown with an intial velocity of 10 m.s-1 at an angle of 60 degrees above the honizontal, then the verticle and horizontal velocities would be as follows;

    [tex]V_{x} = 10\cos 60[/tex]

    [tex]V_{y} = 10\sin 60[/tex]

    Do you follow?

    ~H
     
  9. May 28, 2006 #8
    sort of...so u mean if lets say when the qn gives us the angle and the velocity then i would have to factor it in...otherwise i can ignore the angle in this qn.

    but if there is no vertical velocity...and the horizontal velocity doesnt play a part in getting the rock into the cave, why does the qn still give the horizontal velocity?!
     
  10. May 28, 2006 #9

    Hootenanny

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    RE: AM's post. You need to work out a range of initial velocities. As the intial velocity is horizontal, there is no angle. Consider the horizontal (x) and vertical (y) components seperately.

    Horizontally

    You know that horizontally the rock must travel 20m (the width of the canyon). Ignoring air resistance etc., this velocity is constant, therefore the intial velocity is given by;

    [tex]V_{x} = \frac{20}{t}[/tex]

    Vertically

    You that vertically the shortest distance the rock must fall is 5m (I'm assuming the 15m is measured to the base of the cave as I cannot see your diagram). Ignore air resistance, the only acceleration acting in the y plane is gravity, therefore using kinematic equations the minimum time taken for the rock to fall 5m is given by;

    [tex]t = \sqrt{\frac{2s}{a}}[/tex]

    Now, if you input this time into the equation for horizontal velocity, this will give you the maximum intial velocity.

    Next you will need to work out the maximum time taken for the rock to fall vertically (15m), then input this time into the Vx equation to obtain the minimum intial velocity.

    Do you follow?

    ~H
     
    Last edited: May 28, 2006
  11. May 28, 2006 #10
    diagram ...
     

    Attached Files:

  12. May 28, 2006 #11

    Hootenanny

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    Thank you for the diagram Ukitake, however, it will probably take a while for the mentors to approve this attachment. Do you understand what I was saying in my previous post?

    ~H
     
  13. May 29, 2006 #12
    sorry...my internet was down for a long time

    in light of the diagram i have drawn, does wat u stated still hold true?
     
  14. May 29, 2006 #13

    Hootenanny

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    No problem my friend. I cannot say, because as you can see the attachment is still pending. If it isn't approved soon I will PM one of the mentors to see what can be done.

    ~H
     
  15. May 29, 2006 #14
    thanks :) :)
     
  16. May 29, 2006 #15

    Hootenanny

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    Okay, in light of your diagram the vertical range requires alteration, however the horizontal component still holds (the rock still must travel a horizontal distance of 20m).

    Vertically

    Vertically, the minimum distance the rock must fall to enter the cave is 15m and the maximum distance is 25m. So we can now apply equations of uniform acceleration to find the flight time. The equation required is s = ut + 1/2at2. As the rocks are thrown horizontally the intial vertical velocity is zero, thus the "ut" term drops out leaving s = 1/2at2. Rearranging to find t gives;

    [tex]t = \sqrt{\frac{2s}{a}}[/tex]

    Next you need to substitute in the maximum and minimum values for displacement (s) to obtain the maximum and minimum values for flight time. After obtaining the times you need to substitute them into the equation for horizontal motion to obtain the minimum and maximum intial horizontal velocities. Any velocity which lies between these two values will result in the rock being thrown into the cave.

    Do you follow?

    ~H
     
  17. May 29, 2006 #16
    I don't see any reference to angles in the diagram.
    It's a straightforward question .
    If you still have difficulties, feel free to post them .
     
  18. May 29, 2006 #17
    in order to work out the velocity i need acceleration and since acceleration of horizontal plane is 0. that keeps screwing my answers
     
  19. May 29, 2006 #18
    yea but i have problem with the concepts of kinematics...

    we're taught to treat x component seperately from y component. i am given initial velocity for horizontal component, we know gravity (y component) is 9.8, i can work out the time taken for the rock to fly horizontally but since the cave is located in the y axis i cannot use the time, initial velocity...
     
  20. May 29, 2006 #19

    Hootenanny

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    The acceleration in the horizontal plane is zero.

    However, the accleration in the vertical plane is _____ m.s-2?

    Consider the horizontal and vertical components completely sperately.

    ~H
     
  21. May 29, 2006 #20
    9.8

    but i cant use a vector for vertical plane on the horizontal plane....or can i?
     
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