1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Helicopter Motion Question

  1. Sep 15, 2007 #1

    Vii

    User Avatar

    1. The problem statement, all variables and given/known data

    The height of a helicopter above the ground is given by [tex]h=3.00t^3[/tex], where h is in meters and t is in seconds. After 2.00s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

    2. Relevant equations

    [tex]d=v_1t + \frac{1}{2}at^2[/tex]

    3. The attempt at a solution

    First, I figured out what the height would be at 2.00s by substituting t with 2.00s.

    [tex]h=3.00(2.00)^3[/tex]
    [tex]h=24m[/tex]

    Then I took [tex]d=v_1t + \frac{1}{2}at^2[/tex] and plugged in 24m at d and assumed [tex]v_1[/tex] as 0.

    [tex]24=0t + \frac{1}{2}at^2[/tex]
    [tex]24=\frac{1}{2}at^2[/tex]

    And plugged in 9.8m/s^2 into a.

    [tex]24=\frac{1}{2}9.8t^2[/tex]

    Which gave me t=2.21s

    The answer is wrong and I cannot figure out how to do this problem.

    I think my assumption of [tex]v_1=0[/tex] is wrong. Furthermore, I think my assumption that the helicopter stops ascending at the time the package is dropped is also wrong. I've racked my brain on this question for a good 2 days now and I still can't figure out how to arrive at the answer.
     
  2. jcsd
  3. Sep 15, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    Get the velocity of the helicopter by taking the derivative dh/dt... then plug in the time... and you get v1.
     
  4. Sep 16, 2007 #3

    Vii

    User Avatar

    Oooh, now why didn't I think of that before? Thank you!

    But I've arrived at another problem: how to you tackle the height? Do you just use [tex]3.00t^3[/tex] to replace d, rearrange and get solve the quadratic equation?

    i.e.

    [tex]3.00t^3=v_1t+\frac{1}{2}at^2[/tex]
     
  5. Sep 16, 2007 #4

    learningphysics

    User Avatar
    Homework Helper

    No that's not right.

    The 3.00t^3 only works for the moment the object is released... after that it just falls downwards (it is no longer attached to the helicopter)...

    Use the 3.00t^3 to get the initial conditions... ie the objects initial height and velocity... but after that you don't need it.
     
  6. Sep 16, 2007 #5

    Vii

    User Avatar

    Sorry. I'm afraid I still don't get it. I'm really horrible at Physics...

    So, I get the initial conditions (the height and velocity) and substitute it into the equation and solve for t?
     
  7. Sep 16, 2007 #6

    learningphysics

    User Avatar
    Homework Helper

    Yes, just like in your initial post... everything is right except you assumed v1 was 0... find v1... then do the problem just like in your first post.
     
  8. Sep 16, 2007 #7

    Vii

    User Avatar

    Oh! Thank you so much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?