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Finding the initial velocity of a projectile motion

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A baseball is hit at an angle theta. After a time, it lands 7.5m above the point at which it was hit, the land angle is 28 degree below the horizon. The final velocity is 36m/s2. Find the direction and magnitude of the initial velocity.

    Know: vf = 36m/s2
    theta (final)= 28 degree
    H= 7.5m
    Unknown: Vo and theta(initial)

    2. Relevant equations
    Vxf = Vxi= Vfcos(theta)
    Vyf= Vfsin(theta)
    Vy^2= [V(initial)sin(theta)]^2 - 2g(yf-yi)

    3. The attempt at a solution

    I'm breaking down the initial velocity to its component. First I'm finding the x-component of the initial velocity, which is the same as the final Vx since there is not acceleration in the x-direction:

    Vx (initial) = Vfcos(theta)
    = 36m/s2 * cos(28)
    = 31.79 m/s2

    Then I'm looking for the y-component of the final velocity, which is:
    Vy= Vfsin (theta)
    = 36m/s2 *sin(28)
    = 16.9m/s2

    From there, I find the y-component of the initial velocity:

    Vy^2= [Visin(theta)]^2 - 2g(yf-yi)
    16m/s2 = Vy(initial)^2 - 2g(7.5)
    Vy(initial)^2 = 432.61
    Vy(initial)= 20.8 m/s2

    To find the initial velocity, I used a2+b2 = c2 for the magnitude and tan(theta)= Vi/Vx for the direction, and the final answer I found is 38 m/s2 at the starting angle of 33.19


    ***My concern is that I based my solution on the assumption that the landing angle is 28 and so I use that for all my final thetas. Also, I'm not sure using the last equation, to find the initial Vy is valid. Please let me know where I went wrong in this approach. Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 16, 2009 #2
    When the baseball is hit at [tex]\theta[/tex] degrees to the horizontal, does the baseball reach it's maximum height (7.5 m) and fly down again?

    I am also a bit puzzled about what you mean the 28 degrees is.
     
    Last edited: Mar 16, 2009
  4. Mar 16, 2009 #3

    Mentallic

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    Homework Helper

    You approached the question in the same way that I did. Since the question stated that the landing angle is 28o at 7.5m height (I assumed it hit a tall ceiling), there is nothing wrong with using it to solve the problem. In fact, it's necessary to be used. Also, other assumptions are made such as g=-9.8ms-2, no air resistance and thus Vxi=Vxf

    The only problem I noticed is that you kept saying [itex]v=x.m/s^2[/itex]. Velocity is a speed, not acceleration. [itex]v=x.m/s[/itex] would be correct.
     
  5. Mar 16, 2009 #4
    Yeah, as said, I think the question needs to be clarified somewhat, as I am slightly confused to how you came to get the final velocity and its' components.
     
  6. Mar 16, 2009 #5
    Liberator,

    The 7.5m is not given as the maximum height. It is how far above y=0 it landed. The problem does not state what the maximum height it, it just say that y=7.5 at the landing point. I am uncertain about the 28 degree angle as well. I don't quite understand what "28 degree below the horizon mean," does it mean I use 28 for my angle theta or another number?
     
  7. Mar 16, 2009 #6
    Metallic,

    yea...it's whether I need to use 28 or another angle that I'm concerned about. And thank you regarding the m/s for velocity. I keep missing that =)
     
  8. Mar 17, 2009 #7
    I assume it means that the baseball landed at to the horizontal 28 degrees. Also, know that the 7.5 m is how much higher the ball landed at the end will make it a lot harder.
     
  9. Mar 17, 2009 #8

    Mentallic

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    Homework Helper

    The question is fine the way it is laid out; you just need to get your head around it.

    Think of this: A person hits the baseball at angle [itex]\theta[/itex] with an unknown velocity (but lets assume it's large). Now lets also assume there is a ceiling that is 7.5m above the point where the ball was hit. When the ball hits the ceiling, there is an angle of 28o made with the ceiling and the direction of the ball at the point of contact. The ball hits the ceiling (at that fairly sharp angle) with a velocity 36m/s.

    Thats pretty much the question, but reworded. More clear now? :smile:
     
  10. Mar 17, 2009 #9
    yes, thank you
     
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