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Projectile motion stone problem

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data
    a stone is projected at a cliff of height h with an initial speed of 46.0 m/s directed at an angle θ0 = 65.0° above the horizontal. The stone strikes at A, 5.47 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.


    2. Relevant equations

    v0x= vcostheta
    v0y= vsintheta

    vfy= v0y-gt

    3. The attempt at a solution

    first, i broke the vector up into components
    v0x= 19.44
    v0y= 41.69

    vfy= v0y-gt

    y= -1/2gt^2 +v0yt + y0

    then i tried to find b) the speed of the stone

    i did vfy= v0y-gt
    and got -11.92
    that was wrong

    i'm pretty lost when it comes to the other stuff.. please help?
     
  2. jcsd
  3. Sep 18, 2009 #2
    The speed of the stone consists of both the vertical and horizontal components. After finding vfy, you have to perform a vector addition with vfx.
    As with regards to the height of the cliff, just use a kinematics equation involving time, initial velocity, acceleration and displacement.
    The maximum height of the stone occurs when the stone has instantaneous velocity of zero - can you see why this must be the case?
     
  4. Sep 18, 2009 #3
    how is v0x calculated?

    yes, i understand that the velocity at max height is 0 because it's the peak and the direction of the acceleration is changing
     
  5. Sep 18, 2009 #4

    Redbelly98

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    We're given the initial velocity's magnitude and angle. From that you can use trig to get the x and y components, i.e. v0x and v0y.
     
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