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Free Fall problem

  1. Oct 12, 2008 #1
    Hello friends, I am very new to this forum and I´d like to consult you if i have any questions.
    Currently I am studying basic physics for my ug.
    I´d really apreciate it if you could help me out.

    1. The problem statement, all variables and given/known data
    A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roon one second later. You may ignore air resistance.

    2. Relevant equations
    a) If the height of the building is 20m, what must be the initial speed of the first ball if both hit the ground at the same time.
    b) Consider the same situation but now let the initial speed v0 be given and consider the height h of the building as the unknown.
    What must be the height of the building for both balls to reach the ground at the same time for each of the following value s of v0 (i) 6 m/s (ii) 9.5 m/s.

    3. The attempt at a solution
    I was able to solve a) and i believe that the initail velocity must be 0.2 m/s.

    For the second question I have tried for hours but i cant seem to convince myself that this is possible at all. For instance if the ball 1 is in the air, 6 m, and the next second the second ball is dropped with the accelerration -g. How can the ball which has to travel the extra 6 m hit the ground at the same time ball 2 does?
    I am thoroughly confused. Please help me.
    THanks in advance.
  2. jcsd
  3. Oct 13, 2008 #2


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    Hi tlogan,

    I don't believe this answer is correct. Can you show how you got it?
  4. Oct 13, 2008 #3
    Hello sir,
    all i did was finding the velocity of the second ball.
    i used vx²=v0²+2a(x-x0)

    Then I put:

    Vx² = 2 * -9.8 (-20)
    Vx = 19,8 m/s.

    THen I put it into the equation x = x0+vo*t+1/2*a*t², for t = 1

    20 = Vxo

    And then 20 - 19,8 = 0.2 m/s.

    However now that u say it, its wrong.
    I´ve gone through it but cant really figure out how to tackle this problem. !?!
  5. Oct 13, 2008 #4


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    This is where an error is. The t in this equation is the time of flight of the particle. In other words, the equation answers: if the particle starts at x0 with speed v0, what position x is it at after time t?

    But neither of these balls is in the air a total of 1 second, so you would not set t=1. The difference in their time of flights is what is equation to one second.

    So instead, what I would suggest is to write out your equation:

    x = x0+vo*t+1/2*a*t²

    twice (once for each ball). Since they have different times, you can use t1 for the time variable in one equation and t2 for the other. As soon as you write out how t1 and t2 are related to each other, you can then solve the set of equations for v0. What do you get?
  6. Oct 14, 2008 #5
    Hello sir, I really thank you for helping me out.

    I got:

    sqrt(40/9.8) = 20/Vo

    Then I get Vo = 9.9 m/s.

    I dont really know if this is the right answer. If it isnt could you please elaborate on the suggestion you made? Because I dont really know whether I fully grasped you idea or not.
    Thank you
  7. Oct 14, 2008 #6


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    I don't believe this is correct. What I was talking about in my last post was using the equation:

    x = x0+vo*t+1/2*a*t²

    For the ball that is being thrown up, what is x, x0, and a? So you can write down this equation for ball 1.

    For the ball being dropped, what is x, x0, v0, and t? Write down a separate equation for the second ball.

    What two equations do you get? Using those two equations, you should be able to solve for the v0 in the first equation. Do you get the right answer?
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