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2D Kinematics problem (hill projectile)

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data
    3. A cannon ball is launched from a cliff with an initial velocity of 50 m/s 300 above the horizontal. If it’s range is 300 m.

    a)What is the height of the cliff?
    b) What is the impact velocity of the ball right before it strikes the ground?



    2. Relevant equations
    d=v1t+1/2at^2

    d=V*t

    Vav=V1+v2/2





    3. The attempt at a solution

    How do you find the height and impact velocity?
    Horizontal: Vx=Cos30*50= 43m/s
    dx= 300m

    Vertical: Vy=Sin 30*50=25
    a= -9.8 m/s^2

    I am really stuck!
     
  2. jcsd
  3. Jan 19, 2010 #2

    ideasrule

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    Homework Helper

    You need to derive an equation for the range so that you can plug 300m into it. Try writing out the equations of motion for both the x and y directions. Then solve for the x when y=0 (i.e. when the ball hits the ground).
     
  4. Jan 19, 2010 #3
    "initial velocity of 50 m/s 300 above the horizontal."

    What do you mean by 300 above the horizontal?
     
  5. Jan 19, 2010 #4
    sorry i meant 30 degrees
     
  6. Jan 19, 2010 #5
    let the height of cliff be h
    let the angle be 30 degrees
    range = 50cos30 * time = 300
    time = 4root3 seconds

    initial vertical velocity is vsin30 = 25 m/s
    h = 25 - 1/2gt^2
    assuming g = 10m/s^2
    = - 215 meters
    height is 215 meters

    impact velocity is easiest calculated using energy conservation
    both the equation of motion and energy conservation will lead to the same equation and asnwer
    initial energy = 1/2 m 50^2 + mg 215
    final energy = 1/2mv^2
    v = 84.36 m/s

    I don't know if this is right
     
  7. Jan 19, 2010 #6
    The time calculated seems right.

    h = 25 - 1/2gt^2

    I believe your equation is a little bit wrong.

    Y = Vot + 1/2gt^2

    Does your teacher accept g as 10 m/s^2?? I hate that...
     
  8. Jan 19, 2010 #7
    no sorry i was doing this in my head but should i have acceleration as -9.8 m/s with the negative because my velocity is +

    How would you do this problem?
     
  9. Jan 19, 2010 #8
    I was just pointing out that your Y displacement formula has Vo multiplied by time.

    Y = Vot+1/2gt^2

    and yea, gravity would be negative.
     
  10. Jan 19, 2010 #9
    i just feel very unsure doing this problem and when i do other problems i need to refer to my previous examples, could you give me some tips that you use to solve these easily?
    Thanks
     
  11. Jan 19, 2010 #10
    Whenever you have a 2-d kinematics problem I always separate into 2 columns all the known data. In the x-direction and in the y-direction. And if you ever need to know something in specifically the x or y axis and it seems as though you have insufficient data, then try linking the two by finding time.

    That's just me though. And of course keep in my mind which equations solve for which variables: displacement, initial and final velocity, and time.
     
  12. Jan 19, 2010 #11
    thank you
     
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