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Basic Two Dimensional Problem.

  1. Sep 9, 2009 #1
    Hey, everyone! Glad I found this forum, I plan to be using it a lot to try to expand my Knowledge, I currently taking an AP Physics class in my high school with no prior physics experience. The teacher however is really great! Anyway lets see what we got going here.
    (Note ▲=Delta)
    1. The problem statement, all variables and given/known data

    A ball is thrown horizontally from the roof of a building 58 m tall and lands 50 m from the base. What was the ball's initial speed?

    Vertical Known
    ▲Y=-58
    A Sub Y = -9.8m/s
    t=5.918s
    VnotY=?

    Horizontal Known
    ▲X=50m
    VnotX= ?


    2. Relevant equations
    t=5.918
    I got this because assuming that the distance of 58m was traveled going a speed of -9.8m/s (g) it equals 5.918! (Hopefully)


    3. The attempt at a solution
    Its a long shot but I attempted to compare it to a problem done in school which was essentially the same but, involved an angle of the stone coming off the roof. Also different Knows were given. So this was my go


    &delta Y=(VnotX)(t)² + (.5)(g)(t)² I put in what I know

    -58=(VnotY)(5.91)² + (4.9)(5.91)² Attempted to solve for VnotY

    -58=(VnotY)(34.9281) + (171.14)

    (VnotY)=(34.9281)+(229.14) I added the -58 over

    (VnotY)=264.07 : ( I dont think this is right...

    So that is where I am, I may be completely off...

    Ok! So I looked at the problem again found an Inherent flaw,

    ▲y=(VnotY)(t)+(.5)(g)(t)²

    SO! It turns out (VnotY) is Zero...Making that all Null and void...
    So with some Algebra Magic I get t=3.43! With this I hope to get further...Will keep posted!

    Another Update!!!
    While looking through my book found that while somthing is in free fall it has 0 Acceleration in the X-Axis! So that voids Half of the Equation!! Woo Go Physics! Making it simple and I found that the initial speed was 14.57
     
    Last edited: Sep 9, 2009
  2. jcsd
  3. Sep 10, 2009 #2
    well done!
     
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