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An airplane flying with a velocity

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data
    An airplane is flying with a velocity of 85.0 m/s at an angle of 19.0 degrees above the horizontal. When the plane is a distance 106 m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.


    2. Relevant equations
    How far from the dog will the suitcase land? You can ignore air resistance.

    3. The attempt at a solution
    I attempted to use Y-Yinitial = Voy(t) - 1/2(g)(t^2)
    I solved for t and i got t = 4.651 s.
    That answer was incorrect.

    I then tried using Vx = (85m/s)(cos19) = 80.37
    Vy = (85m/s)(sin19) = 27.67

    I then tried adding them using V = sqrt{(80.1^2) - (27.7^2)} = 75.16 m/s

    I then did R = Vox(t) but the answer I got wasn't correct.
    Help!
     
  2. jcsd
  3. Jan 24, 2009 #2

    LowlyPion

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    Welcome to PF.

    The problem is that the vertical velocity continues to carry it upward before it goes down.

    What if you reduced the problem to throwing a ball straight up at 27 m/s and you are standing on a 105m cliff? Find that time and multiply by the horizontal velocity.
     
  4. Jan 24, 2009 #3
    I'm still not sure how to solve this =(
     
  5. Jan 24, 2009 #4

    LowlyPion

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    You have the right method.

    You got the wrong answer from the quadratic.
     
  6. Jan 24, 2009 #5
    Ohh! I didn't realize my mistake. I used 0 for Voy instead of 85m/s which is the initial velocity.
    I solved the quadratic using the positive value and I got 1.168s. I then multiplied that by 85m/s to find the range and I got 99.28m. Does that sound correct?
     
  7. Jan 24, 2009 #6

    LowlyPion

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    No.

    Your initial vertical velocity is your Voy

    Besides your quadratic should yield t to hit the ground. It can't get to the ground in 1 sec from 100 m.
     
  8. Jan 24, 2009 #7
    I'm sorry, I'm confused again.
     
  9. Jan 24, 2009 #8

    LowlyPion

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    Write out the equation for your quadratic. Let's see where you are going wrong.
     
  10. Jan 24, 2009 #9
    -85 (+ or -) sqrt{(-85^2) - 4(4.9)(-106) all divided by 2(4.9)

    I think that when I added the -85 to the problem I ended up with 1.168

    Now when I subtracted the -85 to the problem I got 18.51s. Which seems more realistic.
     
  11. Jan 24, 2009 #10

    LowlyPion

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    But 85 is the velocity of the plane and it is not going straight up.

    Your vertical component of velocity is 85*Sin19° = 27.67

    That should yield 0 = 105 + 27.67*t -4.9*t2

    That quadratic yields 8.246 s

    I use this on-line calculator btw:
    http://www.math.com/students/calculators/source/quadratic.htm
     
  12. Jan 24, 2009 #11
    Ohhh I understand now. Thank you so much for your help!
     
  13. Jan 24, 2009 #12
    How far from the dog will the suitcase land?

    I'm still having issues finding the distance.
     
  14. Jan 24, 2009 #13

    LowlyPion

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    That's the easy part now.

    You have time to hit the ground.
    You have the horizontal velocity projected along the ground ...

    Speed * time = distance.
     
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