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Another question about projectile motion

  1. Jun 13, 2003 #1
    A stunt man jumps from the top of one building to the top of another building 4.9 m away. after a running start, he leaps at an angle of 16 degrees with respect to the flat roof while traveling at a speed of 5.1 m/s. the other roof is 1.9 shorter than the building from which he jumped. i have to find out his vertical displacement upon reaching the front edge of the lower building with respect to the taller building.

    heres my thoughts on what to do: i can use 16 degrees and 5.1 m/s to find the x and y component of the initial velocity. that's all i've got so far.

    any help would be appreciated
  2. jcsd
  3. Jun 13, 2003 #2
    Right. So what's the problem? Once you've done that, how long does it take him to travel 4.9m horizontally? Then, given his initial VERTICAL velocity, what is his vertical displacement after that length of time?
  4. Jun 13, 2003 #3
    still can't get it

    ok i got that, but where does the 1.9 m come in?
  5. Jun 13, 2003 #4
    Presumably it's there to help you learn to distinguish between relevant and extraneous information. :smile:
  6. Jun 13, 2003 #5
    y(t) = -9.81*t^2 + Voy*t + 1.9m

    The jumpers initial height is 1.9m.
  7. Jun 13, 2003 #6


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    the height you need to find is with respect to the taller building.

    The 1.9m is there because either:

    1) The jump isn't enough to clear the gap.


    2) It's there to trip you up with extraneous information.
  8. Jun 13, 2003 #7
    When I fire from the hip, I tend to shoot myself in the foot.

    Working out the actual numbers, it seems that when he reaches the second building, his vertical displacement
    is -3.49m; i.e. his feet hit the wall 1.59m (3.49-1.9) below the roof. That's where the 1.9m height difference comes in.
  9. Jun 15, 2003 #8
    you could find the solution in the Haliday and Resnick book (plane motion). It is solved there.

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