Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Projectile motion, finding initial velocity

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A rock is thrown from a 1 meter height with a 35 degrees angle with horizontal. At x=64 m, its height is y=29 m. Find it's initial velocity.


    2. Relevant equations

    [tex]x=x_{{0}}+v_{{{\it x0}}}t+ 0.5\,a_{{x}}{t}^{2}[/tex]
    [tex]y=y_{{0}}+v_{{{\it y0}}}t+ 0.5\,a_{{y}}{t}^{2}[/tex]

    3. The attempt at a solution
    Let F = initial velocity
    then Vx0 = F cos 35
    Vy0 = F sin 35
    [tex]x=x_{{0}}+v_{{{\it x0}}}t+ 0.5\,a_{{x}}{t}^{2}[/tex]
    [tex]64=F\cos \left( 35 \right) t[/tex]
    [tex]t=64\,{\frac {1}{F\cos \left( 35 \right) }}[/tex]

    [tex]y=y_{{0}}+v_{{{\it y0}}}t+ 0.5\,a_{{y}}{t}^{2}[/tex]
    [tex]29=1+64\,\tan \left( 35 \right) - 10035.20000\,{\frac {1}{{F}^{2}
    \left( \cos \left( 35 \right) \right) ^{2}}}[/tex]
    solving for F gives 102 m/s
    the answer in the book is 42 m/s


    sorry i am new to latex
     
  2. jcsd
  3. Sep 29, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi fishingspree2,

    The number 10035.2 in the above equation does not look right to me.
     
  4. Sep 29, 2008 #3
    sorry the calculation was made using 20070.4 i don't know how the 10035.2 showed up

    i get 102m/s using 20070.4 and it is not correct according to the book
     
  5. Sep 29, 2008 #4

    alphysicist

    User Avatar
    Homework Helper

    I'm not getting 102m/s using 20070.4; can you show the details about how you get 102m/s?
     
  6. Sep 29, 2008 #5

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I sorted out the constant. Thanks alphysicist.

    It is usually a good idea to keep your expressions symbolic until the very end.

    Show us how you are getting 102m/s.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook