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Calculating average speed.

  1. Sep 12, 2004 #1
    My problem states, that a ball is thrown from the top of a building. At the same instance a person is running on the ground at a distance of 32.6 m away from the building. I have to find the average speed the runner must be traveling to catch the ball at the bottom of the building.

    So far I have calculated everything related to the ball. As in how long it was in the air, it's velocity, and it's maximum height. All calculations I believe are correct.

    Now, however, I'm stuck in that I don't know what to do next. I know I need to find the average speed the runner must be going. So I'm assuming that the velocity of the ball and the time the ball is in the air will play a factor in this. I'm just stuck I don't know where to go from here.
     
  2. jcsd
  3. Sep 12, 2004 #2

    Doc Al

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    Realize that the runner better make it to the building (travel 32.6 m) in exactly the time it takes for the ball to hit the ground. You have the distance; you have the time. Find the speed.
     
  4. Sep 12, 2004 #3
    Just to clarify, time is how long the ball is in the air, correct? So I need to find velocity, and I have distance, and time. Don't I need a third variable to find the velocity?
     
  5. Sep 12, 2004 #4

    Doc Al

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    Right.
    V = D/T.
     
  6. Sep 12, 2004 #5
    Of course. Maybe next time I should just look it up instead of asking a million silly questions.
     
  7. Sep 12, 2004 #6
    Well I must have done something wrong finding the time. Because with everything that I did, I still don't have the right answer.

    Here's the problem word for word and my work that I did:

    A ball is thrown upward from the top of a 24.8m tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 32.6m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building.?

    So to find time I used the equation: T = V - Vo/A And found T to be 1.22 s. I'm thinking now that I did this part wrong.
     
  8. Sep 12, 2004 #7
    Ok well I realized that I didn't double my T answer and thought that solved the problem, but it did not.
     
  9. Sep 13, 2004 #8

    Doc Al

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    finding the time

    Use this kinematic equation for position as a function of time:
    [tex]y = y_0 + v_0 t + 1/2 a t^2[/tex]
    In your case:
    [itex]y_0 = 24.8 m[/itex]
    [itex]y = 0 m[/itex]
    [itex]v_0 = 12 m/s[/itex]
    [itex]a = -9.8 m/s^2[/itex]
    Solve the equation for t.
     
  10. Sep 13, 2004 #9
    Forgive me here but how do I get t alone? Since there is VoT and then 1/2AT^2.
     
  11. Sep 13, 2004 #10

    Doc Al

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    Time for you to get familiar with quadratic equations!
     
  12. Sep 13, 2004 #11
    I suppose it is. LOL
     
  13. Sep 13, 2004 #12
    Ok so I've been trying like a good student should. LOL This is what I've gotten so far. I believe this is my quadratic equation: 0 = 24.8 + 12t - 4.9t^2

    Still I have no idea how to solve for just t? There has to be an easier way to do this.
     
  14. Sep 13, 2004 #13

    Gokul43201

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    No, there isn't. That's a simple quadratic aquation of the form : Ax^2 + Bx + C = 0.

    What are the solutions of such a quadratic ? x = ....... ?
     
  15. Sep 13, 2004 #14

    Gokul43201

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    Actually, there is another way to do this problem that avoids the need to solve the above type of quadratic equation. I'm not going to say what that is, since it would do you a whole lot of good to learn the solution to quadratic equations - it's not hard at all, and will serve you well in the future.
     
  16. Sep 14, 2004 #15

    Doc Al

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  17. Sep 14, 2004 #16
    Apparently, I'm just not supposed to get this one right. I found time using the quadratic equation. Well at least I think I did. However each answer was positive. So I tried to find the velocity using either answer and both were wrong. What is my problem here?

    My quadratic equation was : 0 = 24.8 + 12t - 4.9t^2

    So in solving for t my answers were 0.62 if I added the top two numbers and 1.83 if I subtracted the top numbers. However then when I tried to find the velocity with either number I had no luck. Where am I going wrong?
     
  18. Sep 14, 2004 #17

    Doc Al

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    You must have made an error. One answer is negative, the other positive. Show your work.
    Right.
    I've no idea what you mean by "added the top two numbers". I suspect you are just trying things at random! :grumpy: Don't do that: Solve the quadratic properly!
     
  19. Sep 14, 2004 #18

    Integral

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    If you told us how you arrived at the numbers you give we may be able tell you what you did wrong.

    Those are not the numbers I get as roots to the equation you give.
     
  20. Sep 14, 2004 #19
    I'm not just trying things at random. Contrary to what I might be saying, I actually do need to learn something from this class. I knew what I said was quite vague, but since EVERYONE knows how to do quadratic equations, I naturally assumed that someone would understand. Give me a minute to properly show you my work without it being all jumbled.
     
  21. Sep 14, 2004 #20
    OK. So I must be as dumb as I am protraying myself on here. As I was preparing my work to show you, I noticed that I forgot a negative sign.

    In saying that, I found t to be 3.79. Would I be correct on this?
     
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