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Yet another free fall question.

  1. Sep 13, 2004 #1
    This class is killing me.

    Here's my problem: A woman on a bridge 91.0m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 2.70m more to travel before passing under the bridge. The stone hits the water 4.90m in front of the raft. Find the speed of the raft.

    Ok so the object is to find the speed of the raft. In order to do so, I have to find some variables of the stone.

    I'm guessing that I need to find the velocity, and time the stone is in the air. Is this correct?

    If so, then I have no idea on how to start for either of those. To find time I would need to know the velocity and to find the velocity I would need to know the time. Or at least this is my thinking. A little direction to get me started would be greatly appreciated.
     
  2. jcsd
  3. Sep 13, 2004 #2
    Does the stone hit 4.90m behind the raft? I'm assuming that is so. First you'll want the find how long it takes the stone to fall 91.0m at an acceleration of 9.8m/s^2. It takes the raft that amount of time to move 2.70m plus 4.90m. Velocity is just distance over time. Hope that helps.
     
  4. Sep 13, 2004 #3

    nrqed

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    Yes, the first thing is to figure out the time it takes to the stone to hit the water. Once you know that time, you may use the initial and final positions of the raft to find its speed (I won't detail that part).

    Ok, to figure the time of fall of the stone (forget about the raft for now), you use the equation for the vertical position of an object subjected to the force of gravity only (I am assuming that air friction is negligible). The equation is (sorry, for some weird reasons my equations are not showing up properly.I hope you'll understand the notation):

    y_f = y_i + v_{yi} t - {1 \over 2} g t^2

    where I use the labels "i" and "f" to mean "initial" and "final" (some textbooks use y_f = y and y_i = y_0 ) and g = 9.80 m/s^2.

    Ok. In your case the stone is released from rest, so its initial velocity is zero. For the values of the coordinates, you have to choose an origin. It can be placed anywhere you want (at the level of the water, of the bridge, in between, at 200 m up in the air, whatever). I'll pick the level of water. In that case, y_i = 91.0 m and y_f =0 . Now you can find the time of fall of the stone. And then you can turn your attention back to the stone and find its speed.

    Good luck.

    Pat
     
  5. Sep 13, 2004 #4

    Gokul43201

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    When you say 'velocity' what velocity are you refering to ? The velocity of the stone continuously increases (downwards) after the drop. That's what acceleration means.

    Anyway, you still should learn the solution to quadratic equations, before doing this. I find it hard to believe that quadratic equations have not been covered before this, or have never come up before this problem.

    To ease the pain : If, Ax^2 + Bx + C = 0, the solutions are

    x = {-B + SQRT(B^2 - 4AC) } / 2A and

    x = {-B - SQRT(B^2 - 4AC) } / 2A.

    Remember, if you're solving for t, only a positive solution is realistic.

    I strongly suggest you go over the equations of motion, and look at some solved examples to see how they apply. Surely your text will have this.
     
  6. Sep 13, 2004 #5
    Gokul43201 ~ I haven't covered quadratic equations since high school and that was over 6 years ago. I haven't taken any math classes at all yet while in college and this is my first physics class. So yeah, you could say I'm a bit rusty. I can't devote all my time on one problem, I have to move on to the next and at least get it started. This class is rediculously hard for me, and there are a rediculous amount of problems I have to do per week. So if one problem isn't working out for me in the beginning, I move on to another problem that I might be able to manage and then later come back to the one I'm stuck on. It may not be the right way of doing things, but it's what's working for me now. Some points are better than no points at all. Thank you for your help, however, it is greatly appreciated. After doing some searching on the net, I found the equation you gave. I'll have to work at that later.
     
  7. Sep 13, 2004 #6
    Uhhhmm...there is no need for the quadratic equation on this problem. It can be solved by using the general kinematic equation in most Physics textbooks.
    Since the stone is at rest and has no initial velocity....we can use:

    distance = (1/2)at^2

    We know the distance the stone travels, which is -91 (since its downward, its negative). We know the acceleration, which is -9.8 m/s^2 (since its downward, its also negative) With this info, you can solve for t.

    Now that you know, t, since:

    distance = velocity * time

    Find the distance the raft traveled...and you know the time...you should be able to get velocity.
     
  8. Sep 14, 2004 #7

    Gokul43201

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    Ummm... is this not a quadratic ?

    I think it's important for BlackMamba to at least memorize the solutions of a quadratic, because you will never get through a kinematics course without that knowledge. Ignoring it now, so you can move on will only get you stuck further down, again and again.

    I mean this in your best interests, BM.
     
  9. Sep 14, 2004 #8
    Ok so I calculated the time. And using the initial quadratic equation, so Gokul43201, I should be set in this area.

    As far as finding the rafts distance again, I'm stumped. The raft has 7.20m more to travel before passing under the bridge but the stone hits the water 4.90 m in front of the raft. Oh.......... I think I see. I should just subtract 7.20 and 4.90 and that is the distance the raft travelled. Correct?

    If so then I know how to find the velocity.
     
  10. Sep 14, 2004 #9

    Gokul43201

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    Yes, if this is really 7.2 m, the difference gives you how much the raft travelled during this time.

    NOTE : In your original post, you wrote 2.7 m, and that got some people (including me) quite confused. 7.2 makes more sense.
     
  11. Sep 14, 2004 #10
    Oh, I didn't realize I wrote that backwards. Sorry about that. But yes it is really 7.2 not 2.7.

    Thanks again for everyones help! I really do appreciate it. :)
     
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