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Kinematics in two dimensions

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data

    An airplane with a speed of 59.6 m/s is climbing upward at an angle of 40° counterclockwise from the positive x axis. When the plane's altitude is 600 m the pilot releases a package.
    (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.

    (b) Determine the angle of the velocity vector of the package just before impact.

    2. Relevant equations

    Eight kinematics equations

    3. The attempt at a solution

    completely stuck.. i don't even know where to begin
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 2, 2009 #2
    Welcome to PF kcalhoun!

    This question requires you to split the problem into velocity components and then use your constant acceleration formulae. Also think about what the package does once it is released. So to start with, can you tell us what the speed is of the plane in the horizontal (positive x) and vertical directions (positive y)?

    The Bob
     
  4. Mar 2, 2009 #3
    for the horizontal it would be 59.6cos40
    and the vertical would be 59.6sin40
     
  5. Mar 2, 2009 #4
    Nice. So, if we can find out how long the package takes to fall to the ground, then it would seem logical that we can find out how far it travels due to time. So what do we now need to consider?

    The Bob
     
  6. Mar 2, 2009 #5
    solve for t in one of the constant acceleration equations?
     
  7. Mar 2, 2009 #6
    Yes, in short. However, remember to consider the conditions the package is under once released. What is it's inital vertical velocity? etc. The key is what is positive and what is negative.

    The Bob
     
  8. Mar 2, 2009 #7
    ok so using the equation vy=v0y+ayt i get.. 0=59.6sin40+(-9.80)t and t = 3.909?
     
  9. Mar 2, 2009 #8
    That wasn't the one I was thinking of. What you've worked out is the time it takes for the package to come to rest whilst in mid-air.

    The Bob
     
  10. Mar 2, 2009 #9
    I'm not sure what you are looking for then...
     
  11. Mar 2, 2009 #10
    Okay. Once the package is released, it continues to travel upwards as it still has a vertical component. Gravity takes effect and slows it down such that its velocity after 3.909s is 0ms-1. So, you can either start over with one equation that will do it all for you (e.g. to get total time) or you can stick with this and you can now work out how far it has travelled vertical in addition to the 600m and thus the total time for it to be released and fall to the ground.

    Do you understand why what you've worked out is the time for it to be at v = 0 in mid-air?

    The Bob
     
  12. Mar 2, 2009 #11
    not really no but, was this the equation that you were thinking of... y=v0y+1/2ayt^2?
     
  13. Mar 2, 2009 #12
    That was the one I was thinking of but I'd rather you understood why. The plane ascends and the package is released. The package has the same vertical component as the plane and it is immediately constant, e.g. no external force, after released. However, gravity now acts on the package. (At this point I should note that I will assume negative y is actually positive and positive y is negative but it doesn't really matter that much.) So the package is travelling at -59.6sin(40). This is our initial velocity. We also know at what height above the ground the package was released (600m) and we know that a = g = 9.8ms-2. Does that help you see what I'm getting at? Also, the reason t = 3.909s is write for what I said is because the formula automatically works out the next time something happens (in this case when v = 0).

    The Bob
     
  14. Mar 2, 2009 #13
    ok yeah that makes sense so now when i solve for t i get 114.631 then i need to find delta x to solve part a?
     
  15. Mar 2, 2009 #14
    The equation is:
    s = u*t + 0.5*a*t^2
    where s is the displacement, u is the intial velocity, t is time and a is the constant acceleration. Your equation in post 11 had a missing t, my bad... sorry. I'm not doing very well here.

    The Bob
     
  16. Mar 2, 2009 #15
    right i used that equation to solve for t and i got 600=59.6sin40t+1/2(-9.8)t^2 which yields 114.631 right? then where do i go from there?
     
  17. Mar 2, 2009 #16
    Can you put your working out up please? What you have will yield an imaginary number.

    The Bob
     
  18. Mar 2, 2009 #17
    (600-59.6sin40)/(1/2*9.8)=114.631 does that look right?
     
  19. Mar 2, 2009 #18
    Where have your 't's gone????? You can't solve for t without there being any in the equation. This is a quadratic equation so you're need the quadratic formula.

    The Bob
     
  20. Mar 2, 2009 #19
    what i did was take the original equation y=v0yt+1/2ayt^2 and divided by t to get y=v0y+1/2ayt and the i solved for t to get t=(y-v0y)/(1/2ay) and then i plugged in the numbers.. t=(600-59.6sin40)/(1/2*9.8) so t=114.631 is that not right?
     
  21. Mar 2, 2009 #20
    It's a good idea but it will not work here because 't' could be positive or negative. You can only divide if you know it is always positive or always negative. So now try it as a quadratic.

    The Bob

    P.S. Sometimes it's worth thinking logically. It only takes an object, initially at rest, just over 11 seconds to fall 600 metres so 114.631 seconds should stand out as not being right.
     
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