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: Kinematics Question (two objects thrown up)

  1. May 8, 2007 #1
    I'm sure this is an easy question but I just can't seem to understand wich equations I need to use. A ball is thrown vertically upward with an initial speed of 11m/s. One second later, a stone is thrown veritcally upward with an initial speed of 25 m/s. Find the time it takes the stone to to catch up with the ball.

    So right now I have for the ball: a= -9.80 m/s^2 initial V=11m/s initial y=0 t=t2-1 and for the stone a=-9.80m/s^2 initial V=25m/s initial y=0 t=?

    So the way I was thinking is that the question is asking at what time the displacements are equal so I thought the equation I am supposed to use is (V^2-iV^2)/2a=(V^2-iV^2)/2a and solve for the V. This doesn't seem to work. Am I on the right track or just totally way off base? Any help or hints would be appreciated. Thanks.
     
  2. jcsd
  3. May 8, 2007 #2
    Considering that you have initial velocities, accelerations and a relation between time taken, how about the second equation of motion?
     
  4. May 8, 2007 #3
    the movement 2 events equivalent displacement

    s1 = u1t1 + (1/2)gt12
    s2 = u2t2 + (1/2)gt22

    s1 = s2

    u1t1 + (1/2)gt12 = u2t2 + (1/2)gt22



    but substitute t1 by t2+1
    (find t2) time of the second ball moving
     
  5. May 8, 2007 #4
    Could you tell me why it is t2+1 instead of t1+1? I'm just thinking that because the stone is thrown up one second later. As in ball=t stone=t+1 ? are you saying it's the other way around?
     
  6. May 8, 2007 #5
    t2=stone's time
    t1=ball's time

    Because however long the stone has been flying (t2), the ball has been flying for one second longer (t1=t2+1).
    If you prefer, you could also say that no matter how long the ball has been in the air (t1), the stone has been in the air one second less (t2=t1-1)
     
  7. May 9, 2007 #6
    I get a negative value for t if I do this and a positive value the other way.. is my algebra just off? Thanks for the help.
     
  8. May 9, 2007 #7
    I would assume that yes, your algebra is off. But if you show what you did, you'll probably get a more precise and accurate answer.
     
  9. Aug 2, 2008 #8
    so much for getting an answer on this stuff eh? just a bunch of posts to show how great they are at physics and that you arent
     
  10. Nov 8, 2010 #9
    I'm still having problems with this question where do i go from

    u1=11m/s
    u2=25m/s
    s1=s2
    t1=(t2+1)

    u1t1 + (1/2)gt12 = u2t2 + (1/2)gt22

    (11m/s)(t2+1)+(4.9m/s2)(t2+1)2=(25m/s)(t2+1)+(4.9m/s2)(t2+1)2
     
  11. Nov 8, 2010 #10
    math is wrong you only insert the (t2+1) onto one side of the equation..
    But (t1-1) works better also our g is - because we are throwing the ball against gravity

    (11m/s)(t1)+(-4.9m/s2)(t1)2=(25m/s)(t1-1)+(-4.9m/s2)(t1-1)2
    This is now corrected^^^
    and you just FOIL out

    11t+(-4.9t2)=25t-25+(-4.9t2)+9.8t+4.9
    cancel (-4.9t2) because its on both sides and simplify for t
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook