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Projectile Motion Leap Question

  1. Sep 12, 2004 #1
    QUick Question:
    A tiger leaps horizontally from a 6.5 m high rock with a speed of 3.5 m/s. How far from the base of the rock will the tiger land?

    I started this problem by trying to find the vert. and horz. components, but I was not given an angle to use sin and cos with......how do i start this problem? :confused:
     
  2. jcsd
  3. Sep 12, 2004 #2

    Pyrrhus

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    Draw a picture, something that will help you.
     
  4. Sep 12, 2004 #3
    i did that already...and it didnt help
     
  5. Sep 12, 2004 #4
    would the angle be 90 degrees since the tiger is jumping horizontally?
     
  6. Sep 12, 2004 #5

    Pyrrhus

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    no, it will be 0 degrees, of course i'm assuming max height is his initial position, it seems like that to me.
     
  7. Sep 12, 2004 #6
    thanks.
    i found how many seconds it will take (1.15)
    but i still dont know how to find out the distance
    sorry for being dumb =)
     
  8. Sep 12, 2004 #7

    Pyrrhus

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    Well at the beginning he only has a Vx component, and remember it's always constant from the beginning to the end, so why don't you use it to calculate the distance?
     
  9. Sep 12, 2004 #8
    would the final answer be V(initial)*t = 3.5 m/s (1.15s) = 4.025 m....
    so 4 meters away??
     
  10. Sep 12, 2004 #9

    Pyrrhus

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    If that's the time when it lands, then yes.
     
  11. Sep 12, 2004 #10
    Thanks for your help...I just found this site tonite and signed up and it is real helpful. Thanks some much for the help, I'm sure I will continue to visit it again. Thanks
     
  12. Sep 12, 2004 #11

    Pyrrhus

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    I've too found this site recently :smile: and yes i believe it is a great resource for students and teachers alike, it's always great to be of help.
     
  13. Sep 12, 2004 #12
    okay I have one more question: Romeo is chunking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with only a horizontal component velocity. He is standing at the edge of a rose garden 4.5 m below her window and 5.0 m from the base of the wall. How fast are the pebbles going when they hit her window? I worked it out and got 5.6 m/s.......is that what the answer is if anyone out there wants to work it out? thanks.
     
  14. Sep 12, 2004 #13

    sorry to go back to a question you already answered, but how did you find the time it will take?

    i found that
    V(0)x = 3.5m/s
    V(0)y = 0 m/s

    would you use this to find the time?
    x(0) would be 4.5 since it's the initial height right?
    x(t) = x(0) + v(0)t + 1/2(a)(t)^2?
    0 = 4.5 + 3.5t + 1/2(-9.8)(t)^2

    is that how you solved for t?
     
  15. Sep 12, 2004 #14
    i said that D=1/2(9.8)(1.15) which equals 5.63
    to find t i did t =square root of 2d/g or 2(5.63)/9.8 which equals 1.15
     
  16. Sep 13, 2004 #15

    Pyrrhus

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    did you finish it, umkat?
     
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