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Projectiles launched at an angle

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data

    A baseball is hit over a 7.5-m-high fence 95 m from the home plate. What was the minimum speed of the ball when it left the bat? Assume the ball was hit 1 m above the ground and its path initially made a 38 degree angle with the ground.

    2. Relevant equations

    y=v(initial, y direction)t-1/2at^2 x=v(initial, x direction)t

    3. The attempt at a solution

    t=95/v 95/v(cos38) ? plug 95/v(cos38) into the y equation? and substitute v(sin38) for initial y velocity? This is where I get lost; our teacher wrote all the work on the board very quickly and came up with an answer somehow, but when I did the work I didnt even come close.
     
  2. jcsd
  3. Dec 16, 2008 #2

    G01

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    You have the following (correct) equation for the height of the ball right when it gets to the fence:

    [tex]y=v\sin(38^o)-\frac{1}{2}g\left(\frac{95}{v\cos(38^o)}\right)^2[/tex]

    Now how high do you want the ball to be so that it just barely the fence? Plug that in for y and when you solve for v, you'll get the minimum speed.
     
  4. Dec 16, 2008 #3
    Ok but I dont understand how to get v by itself, with the cos38 and sin38 there. And in the equation you wrote is "t" supposed to be in there at all?
     
  5. Dec 16, 2008 #4

    hage567

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    There is a t missing in the vsin(38) term in that equation. G01 has substituted for t since you can express it in terms of the horizontal distance and horizontal velocity. As already pointed out, you can figure out what y should be from the information given in the problem. So the equation should look like this:

    [tex]y=v\sin(38^o)\left(\frac{95}{vcos(38^o)}\right)-\frac{1}{2}g\left(\frac{95}{v\cos(38^o)}\right)^2[/tex]

    Now it is just algebra. To isolate v, start by getting the term containing v on its own on one side of the equation (note the v's cancel in the first term!)
     
  6. Dec 16, 2008 #5
    Ok so [95/v(cos38)]^2=[6.5-(tan38)(95)]/-4.9? because sinX/cosX=tanX
     
  7. Dec 16, 2008 #6

    hage567

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    Looks good so far. What would you do next?
     
  8. Dec 16, 2008 #7
    I solved the right side and got 13.8, which i took the square root of (since the left side was squared) and got 3.7. Then I divided that by 95 and got 25.6. So now v(cos38)=25.6? If it does then I could just say 25.6/cos38=v, so 25.6/.79=32.4?
    I might've rounded too much so it might not be exact, and Im not sure if you can have cosX or sinX by itself?
     
  9. Dec 17, 2008 #8

    hage567

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    That answer looks OK to me.
     
  10. Dec 18, 2008 #9
    It was thank you! I appreciate the help!
     
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