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Projectile Motion from a given height

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A baseball is bunted down the third base line leaving the bat from a height of .900m with initial v = 3.00 m/s at angle 20° above the horizontal.
    I calculated that
    x(range)= 1.72m
    t(total)=.61s
    vx=2.82 m/s
    vy=1.03 m/s
    -What is the speed of the ball before it hits the ground?
    -What is the max height of the baseball above the horizontal in this travel?
    -What is the ball's centripetal and tangential acceleration at max height?
    -What is the ball's centripetal and tangential acceleration at the instant just before the ball hits the ground?
    -What are the radiuses of the baseball path curvature at max height and at the point of grounding? (What does this question even mean?)

    2. Relevant equations
    (y-y0)=V0yt+at2/2
    v=v0+at
    ac=v2/R
    at=dv/dt


    3. The attempt at a solution
    For calculating the max height, I keep getting y max as equal to .478 m, which makes no sense since its even less than the initial height... Anybody know what I am doing wrong?
    Additionally for for the speed before it hits the ground I'm pretty sure that means normal velocity, yet when I plug in the numbers to the equation I get a velocity of -2.98 m/s which really does not make sense.
    Regarding the centripetal acceleration, I just wanted to clarify if r(radius) would be equal to half of the distance traveled or not...
    And how to calculate tangential acceleration? I have no clue...
    Also what would the difference between the acceleration at max height and before it lands?
     
  2. jcsd
  3. Feb 7, 2013 #2

    haruspex

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    Much more, surely.
    Do you know about curvature as a local property of a curve? Just as you can have a tangent line at a point on a curve, you can also have a tangent circle, which fits both the slope and curvature of the curve at that point.
    Without seeing your working?
     
  4. Feb 7, 2013 #3
    Well to get the y max I first calculated time, which would be equal to -2v0/g, which got me .61 s. I then multiplied vxt which got me 1.72. How is that wrong?
     
  5. Feb 7, 2013 #4

    haruspex

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    That's the time to do what, exactly? Remember that the ball is struck while 0.9m off the ground. And what velocity are you using in that equation?
     
  6. Feb 7, 2013 #5
    The time it would take the ball to travel, but I realized the error in calculating time. I now used the equation y0-y=v0(sinβ)t-.5(g)t2 and got .55 s for t. To get range, I then multiplied the time by vxt=3cos20(.55) and got 1.55 m. Sound about right? I am now trying to calculate the speed of the ball right before it hits the ground using equations like v=v0+at and get completely different answers. Is the speed calculated supposed to be in a specific dimension or something?
     
  7. Feb 7, 2013 #6

    haruspex

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    Yes. (Seemed too low until I realised that 3m/s would be a wet kipper of a bunt.)
    Again, show your working. (You will need to calculate the x and y components separately, then combine them into a single speed.)
     
  8. Feb 7, 2013 #7
    well i got the velocity x component by mulitplying initial velocity by cos(20) and got the y component by multiplying initial velocity by sin20 and then subtracting it by g(time). wait.... do I take the sqrt of these added and squared? just remembered..... so sqrt (vx2+vy2) right?
     
  9. Feb 7, 2013 #8

    haruspex

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    Yes.
     
  10. Feb 7, 2013 #9
    So I've been trying to calculate the y max height. I thought that would occur at .5t, which would give me .275 s. I then plugged this into the equation y-y0=(v0sinβ)t -.5gt2 so y-y0=-.088 and then added .9 which is y0. But this yields a number that is lower than the initial height, so makes no sense. NExt, ive tried that max height would occur at the time when vy=0, so I used the equation vy=v0sinβ-gt and set vy equal to zero to determine the time when the max was reached. For time I then got .1047 s. I then used the equation y-y0=(v0sinβ)t -.5gt2 again and got .9537 m. Does this sound about right? I'm not quite sure as it seems very small.
     
    Last edited: Feb 7, 2013
  11. Feb 7, 2013 #10

    haruspex

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    You meant .55s/2, I assume. But .55s was the time to hit the ground, not the time to return to the initial height. Anyway, there's a simple way to get max height. It's the point where vertical velocity is zero. Do you have an equation like v2-u2=2as?
     
  12. Feb 7, 2013 #11
    um no I don't have that equation but thats what I was sorta talking about in the second part of my reply
    "I've tried that max height would occur at the time when vy=0, so I used the equation vy=v0sinβ-gt and set vy equal to zero to determine the time when the max was reached. For time I then got .1047 s. I then used the equation y-y0=(v0sinβ)t -.5gt2 again and got .9537 m. Does this sound about right? I'm not quite sure as it seems very small."
     
  13. Feb 7, 2013 #12

    haruspex

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    Yes, that's about right. The equation I referred to is quite useful. It's basically conservation of energy with the mass term cancelled out: mv2/2 = mu2/s + mas, where u = initial speed and v = final speed, s = distance.
     
  14. Feb 7, 2013 #13
    Thanks I have a question regarding centripetal motion.... I know acp=v2/R. So to to find it during ymax, I first started by finding the velocity it would be, which would be the horizontal component since the vertical component is 0, which got me 2.819 m/s (by solving vx=v0cosβ. But now I have no idea how to calculate the radius. Any suggestions sir/mam?
     
  15. Feb 7, 2013 #14

    haruspex

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    You know the velocity and the acceleration, so you can calculate the radius using you formula.
     
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