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Aussie NSW Year 12 Projectile Motion Question.

  1. Mar 6, 2007 #1
    1. The problem statement, all variables and given/known data

    A projectile has a time of flight 7.5s and a range 1200m
    calculate
    (a) its horizontal velocity
    (b) Its maximum height
    (c) the velocity witch which it is produced

    I take ay as -9.8ms-2

    2. Relevant equations

    r = uxt
    dy = uyt + (ayt2/2)
    vy2 = uy2 + 2ah
    ux2 + uy2 = u
    v = u + at

    3. The attempt at a solution

    I hope I am the correct

    a) r = uxt

    ux = r/t
    ux = 1200/7.5
    ux = 160m

    .: Initial Horizontal Velocity is 160m

    b)

    dy = uyt + (at2/2)

    dy = 0 (Total vertical displacement is zero)
    0 = 7.5uy - 275.625
    7.5uy = 275.625
    uy = 36.75

    vy2 = uy2 + 2ah

    vy = 0
    0 = 1350.5625 - 19.6h
    19.6h = 1350.5625
    h = 68.90625 ...

    .: Maximum height is 68.91m-1

    c)

    ux2 + uy2 = u

    (160)2 + (36.75)2 = 26950.5625
    u = 164.1662648 ...

    .: Initial Velocity is 164.17ms-1

    v = u + at

    v = (164.1662648) + (-9.8)(7.5)
    v = 90.6662648 ...

    .: Final Velocity is 90.67ms-1


    Ya ?
     
  2. jcsd
  3. Mar 6, 2007 #2

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    a, b, and c look correct numerically, but your units generally are not. Then you added a part d which is not correct. You've got to look in the vertical and horizontal direction separately. Since energy is conserved, what might the final velocity be?
     
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