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Simple Kinematics question

  1. Jan 9, 2015 #1
    • This was originally posted in a non-homework forum so it does not use the template.
    A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.998 s. How much time passes during his entire trip from the top down to the water? Ignore air resistance.

    Attempt:

    t = 0.998 approx 1 sec
    vo = 0 m/s (v initial)
    g = -9.8 m/s^2

    I solved for vf ( v final) using vf = -g * t = -9.8 m/s. This is also the initial velocity of falling the second half.

    Now using v_f, v_o, and g, I solved for displacement. using vf^2 - vo^2 = 2a*Delta x

    Thus the half the distance would be 4.9 meters. Now I solved for the time to fall the second half of the height using
    vo = =-9.8 m/s , a = -9.8, h = 4.9 meters.

    -4.9 = - 9.8t - 4.9t^2

    t^2 + 2t = 1

    t^2 + 2t - 1 = 1 + 1

    (t-1)^2 = 2 so t = sqrt(2) - 1 or 0.41 seconds.

    For t_total I did 0.41 + 1 second from the given to get 1.41 seconds.


    I'm not sure if this is correct when I searched on yahoo answers, the answer reported was 1.13, but my work seems to make sense. If anyone can please elaborate, I'd appreciate
     
  2. jcsd
  3. Jan 9, 2015 #2

    Doc Al

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    Staff: Mentor

    I'd say your answer is correct.
     
  4. Jan 9, 2015 #3

    Doc Al

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    Staff: Mentor

    To convince yourself of your answer, solve it a different way. For example, since you know half the distance is 4.9, the total distance is 9.8 m. How long does it take something to fall that distance?
     
  5. Jan 9, 2015 #4

    PeroK

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    Alternatively, you have the equation for displacement in terms of time:

    ##s = 0.5at^2##

    Why not plug your times into that and see what comes out?

    Also, you might like to think about the general problem: If something takes t seconds to fall a distance h, how long does it take to fall 2h? Starting from rest, of course.
     
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