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

Kinematics simultaneous motion problem!

  1. Sep 26, 2006 #1
    the top question reads

    "You are on top of a building 186 meters high you throw a ball downard at a speed of 18m/sec"

    After calculating time for hitting the ground final velocity and average velocity the next question says:

    "Suppose instead that at the same moment that this ball was thrown downward from the roof a second ball was thrown upward from the ground with a speed of 32m/s"

    I know that when theres simultaneous motion you are supposed to find what the two objects have in common but I can't seem to find it.

    I found the time to hit the ground for the first ball was 4.59, its final velocity was -62.98m/s and its average velocity was -40.52m/s.

    I figured the time it takes the ball from the ground to reach a velocity of 0 is 3.27seconds and the height it reaches is 52.32 meters I'm not sure what to do now though, somebody please give me a hint or steer me in the right direction. Thank you.
  2. jcsd
  3. Sep 26, 2006 #2
    for the ball thrown downward ...
    h = 186 - 18t - 4.9t^2

    for the ball thrown upward ...
    h = 32t - 4.9t^2

    they meet when h = h ...

    186 - 18t - 4.9t^2 = 32t - 4.9t^2

    solve for t (very easy), then find h.
  4. Sep 26, 2006 #3

    Okay, I understand that they both will hit when h = h, but where are you getting those two equations from, and I calculated a time of 3.72 seconds for them to collide, and a height of .85 meters please let me know if this is right.

    Can anyone let me know if these answers are correct?
    Last edited: Sep 26, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook