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Homework Help: A typical motion question (am I correct?)

  1. Feb 11, 2006 #1
    Hey, I was wondering if someone could check my answer for this question:

    A helicopter is rescuing a boy from a flooded street. Hovering 20 m in the air, a cable is lowered to the child. After it is attached, the helicopter begins to accelerate upwards at 0.2 m/s^2. Eight seconds later, the rope snaps and the child plunges to the ground. How fast (in km/h) was the child traveling when he landed?

    Helicopter:

    a = 0.2m/s^2
    t = 8s
    v1 = 0 (I figured since he is hovering)


    Boy:
    v1 =
    d =
    a = 9.8m/s^2

    I started by calculating the overall distance the helicopter traveled; it came out to 6.4 m, which makes it 26.4 m in the air. The boy must then be 6.4 m off the ground, and thus his fall is a total of 6.4 m.

    Then I calculated the final speed of the helicopter. I figured that the final speed of the helicopter should be the initial speed of the boy. This came out to 1.6m/s

    At this point I have the distance of the boy's fall, his acceleration, and his initial velocity (it turned out to be negative, due to direction).

    I subbed in acc, v1, and d into v2^2 = v1^2 + 2ad and got an answer of 11.31 m/s, which ends up being 41 km/h.

    Is my logic correct? Thanks in advance,

    Nicholas
     
  2. jcsd
  3. Feb 11, 2006 #2

    Astronuc

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    Staff Emeritus
    Science Advisor

    What is v1 in the equation - v2^2 = v1^2 + 2ad.

    Your logic regarding the altitude of 6.4 m and the boys velocity 1.6 m/s is correct.

    At the point the cable breaks, the boys begins to decelerate, from 1.6 m/s until he stops, V=0, at which point he achieves maximum altitude. Then the boy accelerates downward to the ground.
     
  4. Feb 12, 2006 #3
    Ack! I forgot about the upward motion. Thanks Astronuc.
     
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