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Horizontal projectile, NOT GIVEN INITIAL VELOCITY

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    A cannon ball is fired horizontally off the top of a cliff that is
    400 m high above the ocean. The ball hits the water 600 m
    away. With what speed does the cannon ball enter the water?



    2. Relevant equations

    I've tried using

    d[itex]_{x}[/itex] = u[itex]_{x}[/itex]*t

    d[itex]_{y}[/itex] = u[itex]_{y}[/itex]*t + 1/2a[itex]_{y}[/itex]*t[itex]^{2}[/itex]


    and also solving for t
    t = [itex]\sqrt{2y/g}[/itex]



    3. The attempt at a solution

    So I found that t = 9 seconds
    and that u[itex]_{x}[/itex] = 66.4

    but i can't seem to figure out how to find final velocity, or speed.
    I know what the correct answer is (111 m/s), I just can't figure out how it was gotten.

    I have an exam in an hour, any help would be greatly appreciated!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 21, 2012 #2

    Doc Al

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    Staff: Mentor

    You found the horizontal component of velocity, now find the vertical component. Once you have both components, you can figure out the total speed.
     
  4. Feb 21, 2012 #3
    u[itex]_{y}[/itex] = 22.6

    so if i do square root of 66.4^2 + 22.6^2 I should get my answer but i get 70.1 instead.

    how do you find the final based on the x and y components?
     
  5. Feb 21, 2012 #4

    Doc Al

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    Staff: Mentor

    How did you get that result?
     
  6. Feb 21, 2012 #5
    t = 9 sec

    so i did the following:

    600 = uy + 1/2*9.8*9^2
    solving for uy I got 22.6
     
  7. Feb 21, 2012 #6
    you want to use the displacement in the y-axis, not the x-axis
     
  8. Feb 21, 2012 #7

    Doc Al

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    Staff: Mentor

    For one thing, you are dealing with the vertical motion, not the horizontal. For another, you'll need a different kinematic equation--one relating velocity and time. (Using the equation you chose will trivially give you a zero initial velocity--which you already should know.)
     
  9. Feb 21, 2012 #8
    ah yes. okay so that means 400 = uy*9 + 1/2 *9.8*9^2
    so uy = .34

    and i guess i did ux wrong, so ux is actually 600 = ux*9, giving ux = 67

    but from here i still don't know how to find the final velocity
     
  10. Feb 21, 2012 #9

    Doc Al

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    Staff: Mentor

    No. Done more accurately, you'll get uy = 0. (9 seconds is just an approximate answer.) Realize that that equation gives the initial velocity, which is not what you want. You should know that the initial vertical velocity is zero. (You used that fact to find the time.)
    No, you were right the first time. (How did you find the time?)

    How does velocity change for accelerated motion? What's the definition of acceleration?
     
  11. Feb 21, 2012 #10
    velocity increases due to acceleration... a = change in velocity/change in time

    so we have a change from zero to 66.4 over the course of 9 seconds
    meaning acceleration is 7.4 m/s ?

    and i know v = vi + at
    so i feel like i could do v= 0 + 7.4*9

    giving me 66.4, which is wrong.
     
  12. Feb 21, 2012 #11

    Doc Al

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    Staff: Mentor

    Good. Now apply that to the vertical motion.

    No, the horizontal motion is not accelerated.

    That's the equation you need. Apply it to the vertical motion.
     
  13. Feb 21, 2012 #12

    Integral

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    Staff Emeritus
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    Gold Member

    What is the acceleration of a falling body?
     
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