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Projectile motion apple throw problem

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data

    A child sitting in a tree throws his apple core from where he is perched (4.0m high) with a velocity of 5.0 m/s [35 degrees above horiztontal], and it hits the ground right next to his friend.

    A) How long is it before the apple core hits the ground?

    B) How far from the base of the tree will the apple core land?

    C) What is the velocity of the apple core on impact?


    2. Relevant equations

    part A) I used dy= Viy x t + 1/2(ay)(t)^2

    part B) I used dx = (Vx)(t)

    part C) I used Vf^2 = Viy^2 + 2ay(dy)

    Then Pythagorean theorem to find out the final velocity and tan function to find the angle


    3. The attempt at a solution

    My answers were as follows:

    A) 1.27 second

    B) 5.2m

    C) 9.9 m/s @ 65.5 degrees below horizontal


    Any feedback is greatly appreciated.
     
  2. jcsd
  3. Aug 10, 2011 #2

    ehild

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    Gold Member

    It is all right, but there are rounding errors. Use more significant digits during calculations.

    ehild
     
  4. Aug 10, 2011 #3
    Your rationale is correct! However, I have a few minor suggestions -

    1. For A), B), and C), I agree with ehild, there are rounding problems. My results for A) are 1.23s (for g=-10) or 1.24s (for g=-9.8).

    2. For C), why not using Vyfinal=Vyinitial-g*t? I think this makes calculation easier, as you avoid doing sq root.

    :)
     
  5. Aug 11, 2011 #4
    Oh good! I know how to do this. Sounds like a problem I had trouble with, but I remember a really good way to figure it out. So, as long as you know that dy=vi,yt + 1/2gt^2 then you could figure out time. since the apple core falls 4 meters below the point of full quadratic motion (basically 4 meters below the y point in which it returns),

    -4meters= vi,yt +1/2gt^2

    vi,y=5m/s(sin35) and plug that in and use the quadratic formula to find time and use common sense to determine which time makes more sense. and use that time to find distance which is easy.

    As for the third question, lets see v^2=v0^2 + 2adx and v=v0 + gt and v^2=v0^2 + 2gdy
    So, on impact, v^2=(5.0m/s)^2 +2g(-4meters)

    Figure out velocity from there :)
     
  6. Aug 11, 2011 #5
    Urgh, How are you guys getting 1.23 I dont know what I am doing wrong??

    for time I use

    4 = 2.86m/s x t + 1/2(9.8 m/s^2) t^2

    I keep getting 1.27s

    Can anyone point out where I am going wrong? This is killing me
     
  7. Aug 11, 2011 #6

    gneill

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    Staff: Mentor

    Can you explain the terms in your equation? How do the signs of the terms in your equation relate to your chosen coordinate system?
     
  8. Aug 11, 2011 #7
    dy =vi,y x t + 1/2 a,y x t^2
     
  9. Aug 11, 2011 #8

    gneill

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    That's not a clear explanation. When you stuck in the numbers you wrote:

    4 = 2.86m/s x t + 1/2(9.8 m/s^2) t^2

    Explain your choice of signs for each of the terms.
     
  10. Aug 11, 2011 #9
    I got the 4 from the height (dy) of the child.

    2.86m/s from 5m/s x Sin35

    t = unsolved for

    9.8 m/s^2 = acceleration due to gravity
     
  11. Aug 11, 2011 #10

    gneill

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    So... The apple core is going to land 4 meters higher than the child in the tree, and is accelerated upwards at a rate of 9.8m/s2?
     
  12. Aug 11, 2011 #11
    No it will land 4 m below the child on the ground, and it is thrown at an angle of 35 degrees at 5m/s
     
  13. Aug 11, 2011 #12

    gneill

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    Does your equation reflect those details?
     
  14. Aug 11, 2011 #13
    I dont think so, should it be -4 m dor dy and -9.8 m/s ^ 2 for ay?
     
  15. Aug 11, 2011 #14

    gneill

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    Yup! :smile:
     
  16. Aug 11, 2011 #15
    That still gives me 1.27s though. I still don't know what I am doing wrong to keep being .3s off the times that others are getting.
     
  17. Aug 11, 2011 #16

    gneill

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    Can you show your calculations?
     
  18. Aug 11, 2011 #17
    My original calculations were:

    4 m = 2.86 m/s x t + 1/2 (-9.8m/s ^2) t^2

    4 m = 1/2(-9.8m/s^2) t^2

    This is in square root, dont know how to do on the forum:

    t = 2(4m) / (9.8m/s^2)

    t = 1.27s
     
  19. Aug 11, 2011 #18

    gneill

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    You've still assumed that the landing point is above the launch point.

    What happened to the velocity term (t) ?
    You need to use the full quadratic formula here. There's a "t" as well as a "t2" term to deal with.
     
  20. Aug 11, 2011 #19
    The first t is negated out because its unknown which takes away the 2.86 m/s right?

    I will redo the equation now with the negatives in what I believe is the way to do it.

    -4m = 2.86 m/s x t + 1/2 (-9.8 m/s ^2) t^2

    -4m = 1/2 (-9.8 m/s ^ 2) t^2

    t^2 = -4m / 1/2(9.8 m/s^2)

    And from square rooting the answer of that I get error on my calculator. :(
     
  21. Aug 11, 2011 #20

    gneill

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    No. The term must stay. The t is the unknown that you're solving for.

    The equation is of the form
    [tex]at^2 + bt + c = 0[/tex]
    solve for t.
     
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