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Tiger leap physics

  1. Oct 10, 2007 #1
    A tiger leaps horizontally from a 12 m high rock with speed of 4.5 m/s. How far from base will she land>

    d=vit + 1/2a(t)(t)



    12m=4.5m/s*t
    t=2.66s

    thats where i got stuck
     
  2. jcsd
  3. Oct 10, 2007 #2

    learningphysics

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    The 12m is vertical displacement not horizontal.

    Split the problem into 2 parts: 1) vertical. 2) horizontal.

    The vertical part:

    Use the equation: d = v1*t + (1/2)at^2, where v1 is the initial vertical velocity. what is a?

    solve for t. Then do the horizontal part...
     
  4. Oct 10, 2007 #3

    Doc Al

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    Staff: Mentor

    Treat horizontal and vertical motion separately. To find the time, use that formula for vertical motion (which is accelerated motion with a = g = 9.8 m/s^2) and solve for t:
    [tex]d = 1/2 a t^2[/tex]

    When you find the time, use a formula for the horizontal motion to find the horizontal distance. The horizontal motion is not accelerated.
     
  5. Oct 10, 2007 #4

    Astronuc

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    The are two parts to this problem.

    One is to determine the time to fall 12 m, and the second part is determine who far the tiger travels horizontally during that time.

    So part 1 - what is the forumla for ball some distance (height) under the influence of gravity. Use that to find T.

    Then the tiger leaps horizontally with speed of 4.5 m/s. How far does the tiger travel in time T with a constant speed of 4.5 m/s.

    Please refer to - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

    Start with 'Freefall'.
     
  6. Oct 10, 2007 #5
    i dont think i did this right but is a .00849?

    i did 12 m=4.5m/s*2.66s+(1/2)at^2
    12=11.97+3.53a
    .03/3.53=3.53a/3.53
    a=.00849
     
    Last edited: Oct 10, 2007
  7. Oct 10, 2007 #6

    Astronuc

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    Please refer to Doc Al's comment. With respect to learningphysics's post, the initial vertical velocity is zero. One the tiger leaps (horizontally), it starts falling.

    One uses the equation of motion in the vertical to find time t or T.

    a = g = 9.8 m/s2

    Please refer to the hyperphysics page.
     
  8. Oct 10, 2007 #7
    what do i use to get the horizontal part
     
  9. Oct 10, 2007 #8
    ok i got that the time is 2.66s
     
  10. Oct 10, 2007 #9

    learningphysics

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    That's not right. Can you show your steps? Did you use the equation Doc Al posted?
     
  11. Oct 10, 2007 #10

    Doc Al

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    How did you arrive at that answer? (Note that that was your original answer, which is still incorrect.)
     
  12. Oct 10, 2007 #11
    i am really stuck. i dont no what to next
     
  13. Oct 10, 2007 #12
    i tried to use the equation
    but this is how i came up with my answer
    d=vit+(1/2)at^2
    12=4.5t+(1/2)0t^2
    12=4.5m/s*t
    t=2.66s
     
  14. Oct 10, 2007 #13

    learningphysics

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    d = v1*t + (1/2)at^2

    v1 = 0, so

    d = (1/2)at^2

    taking down as positive and up as negative:

    -12 = (1/2)(-9.8)t^2

    solve for t.
     
  15. Oct 10, 2007 #14
    but i guess a is 9.81m/s^2
     
  16. Oct 10, 2007 #15
  17. Oct 10, 2007 #16
    so how do i apply that to the horizontal part
     
  18. Oct 10, 2007 #17
    i found the answer! x=0+4.5*1.56
    x=7.02
     
  19. Oct 10, 2007 #18
    i got the answer
    it is 7.02
    x=0+4.5*1.56
    x=7.02m
     
  20. Oct 10, 2007 #19

    learningphysics

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    cool. looks right.
     
  21. Oct 10, 2007 #20
    thank you!
     
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