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Homework Help: Confusing problem

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data
    A cannon fires a 0.626 kg shell with initial
    velocity vi = 9.6 m/s in the direction θ = 50◦
    above the horizontal.

    The shell’s trajectory curves downward be-
    cause of gravity, so at the time t = 0.392 s the
    shell is below the straight line by some verti-
    cal distance Delta h. Your task is to calculate the
    distance Delta h in the absence of air resistance.
    What is Delta h?

    2. Relevant equations

    I have no idea.

    3. The attempt at a solution
    I started out by trying to by using using distance y equals initial distance y plus initial y velocity times time plus one half accel times time squared. I plugged in the numbers for where accel was 9.8 (for the parabola) and accel was 0 (for a straight line).

    My teacher has not taught us this, and I'm very impressed we've gotten this far
  2. jcsd
  3. Oct 16, 2008 #2
    i think you are on the right track...
    i dont quite understand the question but ill try give you a hand

    its projectile motion so u need those equations
    and you are dealing in the y direction only...

    have you broken up the initial velocity?
  4. Oct 16, 2008 #3
    No, why would I need to?
  5. Oct 16, 2008 #4
    Did he/she teach you about finding the components of the initial velocity? (use the angle)
    Or about writing two equations of motion, one for vertical direction (accelerated motion) and one for horizontal direction (uniform speed)?
    For this problem you don't really need to look at the horizontal direction. Write eq of motion for vertical motion, once with accleration and a second time for the hypothetical case when a=0.
  6. Oct 16, 2008 #5
    Yes, I was taught about how to find v in y direction and such. What equation should I use?
  7. Oct 16, 2008 #6
    you will only need Vy it seems here

    y = Vy*(t) - (1/2)gt^2
  8. Oct 16, 2008 #7
    Okay I got 2.13 m. Then would I do what I did before, plug it in with the accel being equal to zero, then subtract the two?
  9. Oct 16, 2008 #8
    this straight line.... if it is the path of the shell with a=0 then yes that is right
  10. Oct 16, 2008 #9
    Can I get an answer to check mine with?
  11. Oct 16, 2008 #10
    okies u gotta post yours first cause im not allowed to give u the answer but :)
  12. Oct 16, 2008 #11
    ah not cool. I got 1.86.

    It was 2.88 (straight line)-1.86 (parabola)
    HOwever I feel I completely screwed up
  13. Oct 16, 2008 #12
    ok how did u get ur answer for the parabola?\
    you had 2.13m above and its changeD?
  14. Oct 16, 2008 #13
    For the parabola, I set everything the same cept for the accel, which became 0, so i basically had dist= vel * time.

    And I dunno where i got 2.88, and now I'm getting 3.63. Confirmation...?
  15. Oct 16, 2008 #14
    y = Vy*(t) - (1/2)gt^2

    u have Vy i take it already
    you know t
    g = 9.81m/s^2
    sub it in
    (that is the parabola equation)

    for the linear one
    g = 0
  16. Oct 16, 2008 #15
    can you confirm my answer?
  17. Oct 16, 2008 #16
    2.88 is when a=0
    2.13 when a=g
  18. Oct 16, 2008 #17
    that is what i got

    so now subtract the linear from the parabola
  19. Oct 16, 2008 #18
    Thank you guys so much
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