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Rearranging the Projectile Motion Base Quadratic Formula for Initial Velocity

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data
    What I'm supposed to do is to rearrange the this formula -16t^2+Vt+h and solve it for V.
    V= Initial Velocity t= time (throwing the ball in a parabolic arc) h= height
    I know what the height is and it is 6ft. I also have the time which is 2.03 seconds.
    How do I rearrange for V? I got an answer but I'm not sure and doubting if it is correct.
    2. Relevant equations


    -16t^2+Vt+6 (if you plug in the height.)

    3. The attempt at a solution

    -16t^2+Vt+6 What I started with

    -16t^2+Vt= -6 Subtracted 6 to the other side

    Vt= -6+16t^2 Added -16t^2 to both sides.

    V= (-6/t)+16t Divided both sides (every term) by t. This is my answer so far.

    I have no idea if I'm correct or not and I have the feeling I'm not. Please help, i'm in US grade 9.
    Last edited: May 23, 2010
  2. jcsd
  3. May 23, 2010 #2


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    Homework Helper

    Is that meant to be the physics equation for motion:

    [tex]h=Vt+1/2gt^2[/tex] ?

    So then re-arranging you have [tex]-1/2gt^2-Vt+h=0[/tex]

    It doesn't seem to be consistent with what you started with. But if we assume you started with teh correct equation, yes, you've done it right.
  4. May 23, 2010 #3
    Thank you. What I was talking about was just simply if I had the correct answer for:
    Taking -16t^2+Vt+h
    and rearranging for V to be by itself on one side of the equation.
  5. May 24, 2010 #4
    -16t^2+Vt+h = 0
    -16t^2+Vt = -h
    Vt = -h + 16t^2
    V = (-h + 16t^2)/t

    your right op, but my way is better
  6. May 24, 2010 #5
    Thank you very much guys!
  7. May 24, 2010 #6


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    How is your way better? It's exactly the same...
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