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Projectile From Cliff

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball is fired from a height of 3.0m above the ground, with a speed of 13m/s and an angle of 22 degrees below horizontal. (a) At what time does the ball hit the ground? (b) What is the impact velocity?

    2. Relevant equations

    vx = v0x + axt
    vy = v0y + ayt
    ∆y= v0∙t+1/2a∙t^2

    vx = v0x + axt
    vy = v0y + ayt

    3. The attempt at a solution

    The part I'm really not sure about is what angle I should derive the components from. Any input would be appreciated!

    ay = -9.8m/s2
    ax = 0m/s2
    v0 = 13m/s
    ∆y = -3.0m

    v0x = ???
    v0y = ???
    t = ???
    ∆x = ???

    v0x = cos 22∙(13m/s) = 12.05m/s
    v0y = sin 22∙(13m/s) = 4.87m/s

    Solve for 't'
    ∆y= v0∙t+1/2a∙t^2
    -3.0m = -13m/s ∙ t + 1/2 ∙ (-9.8m/s2) ∙ (t2)
    t = 0.214s

    I made the velocity negative, indicating its downward path. Corrections are in red.

    Solve for impact velocity.

    vx = v0x + axt
    vx = 12.05m/s + (0m/s2) ∙ (0.214s)
    vx = 12.05m/s

    vy = v0y + ayt
    vy = 4.87m/s + (-9.8m/s2) ∙ (0.214s)
    vy = 2.77m/s

    v = 12.05m/s(x hat) + 2.77m/s(y hat)
    Last edited: Sep 29, 2009
  2. jcsd
  3. Sep 28, 2009 #2


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    Shouldn't the 13 be -13 ?
  4. Sep 28, 2009 #3
    Should it be, since it's directed downward? He gave us this problem, and we've never done one like it.. we have done them where the projectile was shot above the horizon.
  5. Sep 29, 2009 #4
    I did make the correction with -13m/s and re-evaluated my answer. Anyone agree/disagree?
  6. Sep 29, 2009 #5


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    The y component is downward, so it certainly should be negative.
    I haven't checked the calculations.
  7. Sep 29, 2009 #6
    Well, in spite of the calculations does my setup look okay? I can take full responsibility if I can't work the calculator. Just want some reassurance the problem setup/equations are correct.

    I appreciate everyone's help!
  8. Sep 29, 2009 #7
    Should I be using -13m/s in the calculation of v0x and v0y? If I do, I'll get negative components which doesn't make much sense to me.
  9. Sep 29, 2009 #8
    Now that I've recalculated answers, I get a positive y component of the impact velocity vector.. which can't possibly be right. :(
  10. Sep 29, 2009 #9
    Hey chrisfnet,

    You found your initial vxo and vyo correctly, the problem you are making is that you are giving your vyo a negative value when you use y = y0 + vyo + 1/2(a)(t)2 formula.

    If you set the problem up so that up is positive and to the right is positive when projectile is launch at an angle of 22º it will have a positive vyo. Its is not until the projectile reaches its height of its trajectory that it will start having a negative vy. This is just like tossing a ball straight up in the air. Going up it will have a positive velocity and coming down it will have a negative velocity. In this problem the "ball" is moving to the right at the same time.

    The other thing that is odd is how you are finding the final velocity. I used the formula: vf = sqrt((vx)2+(vy02)), which gave me a number that made sense giving how the problem is setup.
  11. Sep 29, 2009 #10


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    librab didn't notice that it is shot BELOW horizontal so the initial vertical velocity is downward.

    In calculating the time of fall, you have an error. You used an initial vertical speed of -13 when it should be -4.87. I realize I'm guilty of not noticing that the magnitude as well as the sign was wrong earlier. Sorry; I was in a hurry to do something else. That is a very common mistake; perhaps one should have a big black marker to wipe out the whole velocity immediately after finding its horizontal and vertical components.

    In the next calc to find the vertical velocity, you again have the error in sign on the initial velocity - should be -4.87 since the negative g indicates you are taking down to be negative.
  12. Sep 29, 2009 #11
    HAHA...how did I miss that hmmmmm.
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