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Projectile motion with differing heights

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A cannon is fired at a target on a cliff that is 400 meters away. The cliff has an elevation of 150 meters relative to the cannon. The initial velocity of the cannonball is 100 m/s. Ignoring wind resistance calculate the two possible angles and their respective times.
    X= 400m
    Y=150m
    V_naught=100m/s
    A= -9.81 m/s^2

    2. Relevant equations

    Time X = Time Y

    X=V_naught*cos(theta)*t
    Y= V_naught*sin(theta)*t - 0.5*a(t)^2


    1/(cos(theta)0^2=1+ (tan(theta))^2



    3. The attempt at a solution
    t= 400/100cos(theta)

    0= -0.5*a(t)^2 + V_naught*sin(theta)*t -150

    0= -0.5*a(400/100cos(theta))^2 + V_naught*sin(theta)*400/100cos(theta) -150

    I get stuck here. I'm not really quite sure what to do next. I'm pretty sure I have to use the quadratic equation, but don't know how.
     
  2. jcsd
  3. Sep 26, 2012 #2

    ehild

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    Hi Hedgehog:smile:, welcome to PF.

    Substitute vnought=100 into your equation and simplify. You also know that a=g, the gravitational acceleration.

    ehild
     
  4. Sep 26, 2012 #3
    Thanks for the quick reply!

    I substituted all my v0=100 and simplified to get this:

    -0.5*-9.81(16/1+(tan(theta))^2) + 400*tan(theta) -150=0

    Edit:
    I'm not sure entirely how I would simplify further.
     
  5. Sep 26, 2012 #4

    ehild

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    Your equation Y= V0*sin(theta)*t - 0.5*a(t)^2 is a bit incorrect, it has to be Y= V0*sin(theta)*t + 0.5*a(t)^2, with a=-9.81 m/s^2.
    Using the parentheses correctly, you get
    -0.5*9.81*16(1/cos2(θ))+ 400*tan(θ) -150=0.
    You wrote among the relevant equations that 1/cos2(θ)=1+tan2(θ). Use it, substitute 1+tan2(θ)for 1/cos2(θ). You get a quadratic equation for tan(θ).

    ehild
     
  6. Sep 26, 2012 #5
    Thank you for your help! I understand what to do now.
     
  7. Sep 27, 2012 #6

    ehild

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    You are welcome:smile:

    ehild
     
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