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One stone dropped one is thrown from a building, they reach the ground at the same ti

  1. Jun 2, 2012 #1
    A stone is dropped from the roof of a building; 1.50 s after that, a second stone is thrown straight down with an initial speed of 24.0 m/s , and the two stones land at the same time.

    How long did it take the first stone to reach the ground?


    V=Vo+aT


    I plugged 1.5 s for time, 9.8 m/s^2 for acceleration, Vo=0 initial speed. Got 14.7 m/s for velocity and plugged it back in the equation for velocity to find the time for the drop.
    I got 4.9 sec.
     
  2. jcsd
  3. Jun 2, 2012 #2
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    The fact that the two stones hit the ground at the same time is important, so you have to use the data from each stone. You never used 24m/s in your work.
     
  4. Jun 2, 2012 #3
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    Am I on the right track with the formula?
     
  5. Jun 2, 2012 #4
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    If you plugged it back in the equation T=1.5 s
    How you get 4.9 sec?
     
  6. Jun 2, 2012 #5
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    I plugged it back in using the velocity I came up with trying to find the time the first stone hit the ground in, but I was told I needed to incorprate the other stone's initial speed into the problem.
     
  7. Jun 2, 2012 #6
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    We have 2 stones means we have 2 equations each for a stone.
    1st stone and 2nd. stone travel the same distance.
    1st. stone starts at 0m/s and 2nd. stone with 24 m/s
    Flight time is different by 1.5 sec.
     
  8. Jun 3, 2012 #7
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    Right, so what equations do I need to be looking at?
     
    Last edited: Jun 3, 2012
  9. Jun 3, 2012 #8
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    The equation for stone number one is just
    d = 1/2at1^2

    Stone 2:
    d = 1/2at2^2 + v0 * t2

    Remember that t2 is t1 - 1.50 and that the displacements of the two stones are equal.
     
  10. Jun 3, 2012 #9
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    Alright thanks for the help. I've been dealing with mastering physics and the program in general is very hard to work with especially if you're no sure how to go about solving the problem. Does anybody have any tips for this system?
     
  11. Jun 3, 2012 #10
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    For all kinematics problems, you can write out your variables (displacement, velocity initial and final, time, and acceleration) for each system (so here for each of the two rocks) and then fill in what you can. So in this problem, you can call Vi for Rock1 0, and Vi for Rock2 24, acceleration for each 9.8, call out that each of their displacements is the same d, and say that t1 = t2 + 1.5.
    Then write out the kinematic equations (I think that there are 4 that we used in my class) and see which ones relate the variables that you need to find out.
     
  12. Jun 3, 2012 #11
    Re: One stone dropped one is thrown from a building, they reach the ground at the sam

    Take the two equations I've quoted, and set them equal to each other. The stones have the same displacement:

    1/2at1^2 = 1/2at2^2 + vi * t2

    Now sub in a = 9.8 m/s2, t2 = t1 - 1.50 s, vi = 24.0 m/s, and leave t1 as the unknown. The squared terms can be subtracted out and you are left with a fairly simple linear equation. Let me know if you need any more help.


     
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