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Problem with finding distance

  1. May 27, 2009 #1
    so i am having issues with finding the distance for this question. here is the question and what i tried to do to solve it.

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

    An airplane with a speed of 17.6 m/s is climbing upward at an angle of 42° counterclockwise from the positive x axis. When the plane's altitude is 840 m the pilot releases a package.

    Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.

    known:
    v (initial) = 17.6 m/s
    direction 42 degrees
    height (horizontal) = -840 m
    a in the y direction: -9.8 m/s^2
    a in the x direction: 0 m/s^2

    unknown:
    v initial in the y or x direction
    time
    distance (vertical)

    2. Relevant equations

    d=v(initial)*t+1/2*a*t^2 in the y direction

    and

    d = (avg velocity)(time) in the x direction

    3. The attempt at a solution

    the question is asking what is the distance from the point of the release of the package to when it hits the ground. i calculated the time it would take for the package to reach the ground and used v initial in the y-direction to be 0 since that is the initial velocity in the y direction when at the max height of the projectile (-840m for the package). i plugged into the equation d=v(initial)*t+1/2*a*t^2 --> -840 = 0*t+1/2*-9.8*t^2 = 13.1 s. THEN i found that v intial in the x direction would be 13.1 m/s using trigonometry (13.1=cos42*17.6m/s). By plugging into d=vt, i get 13.1 m/s * 13/1 s = about 171 m. apparently this is the wrong answer. can anyone help me with this? thanks!
     
  2. jcsd
  3. May 27, 2009 #2
    I think that 42 angle is meant to be the flight path angle, i.e. the angle up from the horizontal (since the airplane is climbing upward). It's weird that they describe that as "counterclockwise", but I can't see what else it could mean.

    If that's the case, then both the airplane and the package are initially moving upwards, i.e. the velocity in the y direction is not zero!
     
  4. May 27, 2009 #3
    thank you for responding belliott4488!
    i did what you said and took v initial in the y direction to not be zero. i found this out by using sin42*17.6 m/s = 11.8 m/s. i found time to be 1.9 s using d=v(initial)*t+1/2*a*t^2 in the y direction. then i 1.9s*13.1m/s = 24.9 m. but the website where i submit my homework is telling me this is wrong too =[ i don't understand, i thought the logic made sense.
     
  5. May 27, 2009 #4
    Check your calculations again maybe - I got a different time. Intuitively, 1.9s is far too short. What does your eqn look like when you try to solve for t? How did you find your values?
     
  6. May 27, 2009 #5
    Hm ... I don't get it. Assuming you did your math correctly, that should be right - your logic is fine.
     
  7. May 27, 2009 #6
    oops! Yeah - what DorianG said - I don't get that 1.9 sec when I solve for t. Walk us through that calculation again.
     
  8. May 27, 2009 #7
    It's going to take a bit more than 2 seconds to fall almost 9 football fields. You are confusing your kinematic equations I think. The position equation you want is

    Y_final = Y_initial +(Viy)t + (-4.9t^2)
     
  9. May 27, 2009 #8
    BTW - does the web site give something like 187.7 m as the answer? If so, then all you need is to correct your calculation of t and you're all set.
     
  10. May 27, 2009 #9
    No, that's what jenador was using (setting Y_initial to zero and Y_final to -840 m). I think she just made a calculational error solving for t.
     
  11. May 27, 2009 #10
    Ahh yes, I see, sorry.
     
  12. May 27, 2009 #11
    using Y= (Viy)t + (-4.9t^2) when i plug in numbers, i get:

    -840 = 11.8*t - 4.9 t^2
    is this correct? and initial velocity is not zero, like belliot said right?
     
  13. May 27, 2009 #12
    Yes, that looks right ... so how are you solving it? It's a quadratic equation, right?
     
  14. May 27, 2009 #13
    Okay so add 840 to both sides, so (-4.9t^2) +(11.8t)+840=0

    and then use the quadratic formula. If you plug in 1.9 s in the left hand side of the equation for time you get 845 which does not equal zero, so you know that your time does not satisfy the equation.
     
  15. May 27, 2009 #14
    omg i was not using the quad equation. thank you for pointing that out to me!!

    since someone said it is the quad equation, i just plugged in using a=-4.9, b=11.8, and c=840. i got t =14.35 s and 187.9 m. and the computer said that was the right answer. thank you all so much!! i feel really dumb for not picking that up.
     
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