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Projectile Motion -

  1. Dec 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A baseball is hit at ground level. The ball is observed to reach its maximum height above ground level 3.0 seconds after being hit. And 2.5 seconds after reaching this maximum height, the ball is observed to barely clear a fence that is 320 ft from where it was hit. How high is the fence? (The acceleration of gravity is -32.2 ft/s^2.)

    2. Relevant equations

    Position and velocity vectors and their x and y components
    h=(v_o^2 sin^2 \Theta_o)/2g

    R=(v_o^2 sin^2 \Theta_o)/g

    3. The attempt at a solution

    I am thinking that I have to relate the x component with the h equation. I'm not sure how to go about it. I don't know the angle, and it seems as if I don't have enough information in order for me to figure out the initial value of theta.
    Last edited: Dec 21, 2008
  2. jcsd
  3. Dec 21, 2008 #2

    Did you trying solving for Vo in terms of theta or vice versa, and then using that to your advantage?
  4. Dec 21, 2008 #3
    I should try solving for one and then use substitution to find the value of one, and then use that to find the other value? Is that what your suggesting?
  5. Dec 21, 2008 #4

    In a rough sense, yeah.

    Think of it this way, say you use this:

    y= h + Vo*sin(theta)*t-(.5)*g*t^2 , you know that at t=3, the ball is at the max height, which means that its velocity in the y direction equals 0. So knowing that, you can get a useful equation from the one i provided by deriving it wrt t (dy/dt). So if you set that equation to 0, you can solve for say Vo if you chose (it will be in terms of one of the other unknowns).

    So you can take that back to the x equation: x=Vo*cos(theta)*t

    You were given some information about its x-coordinate at a certain time, so using that knowledge you are able to solve for one of your unknowns, which will allow you to figure out the question at hand.
  6. Dec 21, 2008 #5
    in the equation you provided above is h the original position or should i plug in the equation for max height?
  7. Dec 21, 2008 #6
    That h is actually the height above the ground that the projectile was launched from. So in your question, it says the baseball was hit from ground level, so in this case h=0.

    But h shouldn't even be a factor here because once you derive the equation with respect to t, that h goes away... after taking the derivative you should have 0= Vo*sin(theta)-g*t

    Which is where you would use the steps i suggested about solving for Vo in terms of g,t, and sin(theta).

    Once you solve for you unknown (you'll find you can solve for theta using the steps i presented), you can go back once again to the original y equation i provided and solve for the final height (The y= h +Vo*sin(theta)*t-(.5)*g*t^2 because you will know the time that the ball is above right above the fence, the angle, the h (=0) and g. So when you solve for y, that is the distance the ball is above the ground (or the fence height). All assuming you found your Vo, which can also be found easily at this point.
    Last edited: Dec 21, 2008
  8. Dec 21, 2008 #7
    i got the angle is approx. 18.186 degrees. is that correct? And can you explain the concept behind why you can take the derivative of that equation?
  9. Dec 21, 2008 #8
    Hmm after doing it i got this:

    After taking deriv. i found V0= (gt/sin(theta)), so plugging that back into x i got:

    x=(gt/sin(theta))*cos(theta)*t ----> x= (cot)*g*t^2 ---> 320= cot(theta)*2*(5.5)^2, which i then found theta= 71.71 degrees.

    The concept behind taking the derivative is this:

    You are given a position function as i presented in the forms x=... and y=... ,
    So if you take the derivative of a position function you get a velocity function, if you take a derivative of a velocity function, you get an acceleration function.

    So by taking the derivative of the y position function, that gives you the y-velocity function (and as we pointed out, the y-velocity when the ball is at the max point is 0, which is why we can set the equation equal to zero).
  10. Dec 21, 2008 #9
    why would it be cot(theta)*2*(5.5)^2. Shouldnt it be cot(theta)*-32.2*(5.5)?
  11. Dec 22, 2008 #10
    Yes, i just made a typo but my theta value still stands, i meant to put 32 ft/s. Also, your (5.5) should be (5.5)^2, when you plug it into the x equation remember you have a t there, and another t in your Vo equation
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