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A stone dropped into a river

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data

    A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone? Give your answer in terms of the given variables and g.

    2. Relevant equations

    h=V_o(t)+0.5gt^2

    3. The attempt at a solution
    I thought that you just had to rearrange the equation so that everything equals the initial velocity, but I'm getting the wrong answer on my homework. This is what I got: V_o=(h-0.5gt^2)/t. I'm really confused, could someone please explain how to work this problem?
     
    Last edited by a moderator: Jan 20, 2014
  2. jcsd
  3. Jan 20, 2014 #2

    berkeman

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    I would use the full form of the y(t) equation (include y_0, Vy_0, and the acceleration term), and write it for both stones separately. Be careful about signs to be consistent. Can you write those two equations for us?
     
  4. Jan 20, 2014 #3
    Is this the right equation?: x-x_o=v_o(t)+0.5gt^2
    I'm sorry, I don't really understand how I would wright those equations. Could you give me a hint?
     
  5. Jan 20, 2014 #4

    berkeman

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    You are on the right track, but since this is vertical motion, I would use y(t) as the main variable, and it is traditional to write the equation slightly differently:

    [tex]y(t) = y_0 + Vy_0*t + \frac{1}{2}a_y*t^2[/tex]

    See the "constant acceleration equations" portion of this page: http://en.wikipedia.org/wiki/Equations_of_motion#Uniform_acceleration

    Again, you have to get the signs right, and you need to set a y=0 point (I would set it at the water's surface, but you could set it up on the bridge and have y going negative down to the water).

    Can you write the 2 equations now for the 2 stones y1(t) and y2(t)?
     
  6. Jan 20, 2014 #5
    Would the equation for the first stone look like this?: y_o=-y+Vy_o(t)+0.5g(t^2)
    I'm still confused... I'm not too good at these constant acceleration problems yet.
     
  7. Jan 20, 2014 #6

    berkeman

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    That's okay, it takes practice.

    Assuming that we take y=0 at the water, and y=h as the height of the bridge, the equation for the dropped stone looks like this:

    [tex]0 = h + 0*t - \frac{1}{2}g*t^2[/tex]

    That is, the final y position is zero, the initial starting y(t=0) = h, there is no initial vertical velocity V_y because it is dropped and not thrown, and the acceleration is due to gravity (g), pointing in the negative y direction. Makes sense?

    Now can you write the equation for the 2nd thrown stone? It will only be a little different. But since both of the stones reach the water at y=0 at the same time, you will have two equations that you can use to solve for the time delta-t between the dropping of the first stone and the throwing down of the second stone.

    The problem statement is using "t" as the difference in time between the dropping and the throwing, which is a little awkward. Usually the variable t is used as the general time variable. Maybe use "T" for the delta-time between the dropping and throwing -- that will help to avoid some confusion.
     
  8. Jan 20, 2014 #7
    Thanks for being so patient with me! So the second stone will have an initial vertical velocity because it was thrown? Would the second equation look like this?: 0=h+V_o(T)-0.5g(T^2)
     
  9. Jan 20, 2014 #8

    berkeman

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    That is very close! But think about how to express the difference in times, between t and t-T...

    In kinematic problems like this, I like to picture the motion of the projectiles versus time. Think about how the situation looks if you were an observer, with the 2nd stone being thrown downward later and catching up to the first. Label the times and positions on the time graphs that you picture in your minds eye, and then translate that into the equations that you write down and solve for the times...
     
  10. Jan 20, 2014 #9
    Is it h=V_o(T-t)+0.5g(T-t)^2?
     
  11. Jan 20, 2014 #10

    berkeman

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    I don't see where you are going with that equation, sorry.

    Try this, call the time for the first stone to travel from y=h to y=0 t1. Call the time for the second stone to travel from y=h to y=0 t2. The difference t1-t2 = T. Stone 1 starts travelling downward at time t=0. Stone 2 starts travelling downward with velocity V0 at time t2. Up is positive and down is negative. Can you write the two equations now?
     
  12. Jan 21, 2014 #11
    I'm sorry, I feel like you've tried every possible way to explain this to me, but I'm still confused.
    Would the first equation look like this?:h=-0.5g(t^2)
    And the second?: h=V_o(t)-0.5g(t^2)
    Aren't there supposed to be delta t's in there somewhere?
     
  13. Jan 21, 2014 #12

    berkeman

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    Yes. But I've changed the time that it takes to drop to the water to t1, and fixed the sign on h.

    Close. I fixed the sign on h and V_o, and changed the time to t2 to make it clearer. Remember, the 2nd stone is going to take less time to get to the water because it was thrown down.

    Also, on both, I've changed your (t) into *t, because when parenthesis are used around a variable, it usually means "a function of", like v(t) is velocity as a function of time.

    The delta-t = t1-t2...
     
  14. Jan 21, 2014 #13
    Ok so then you would just solve the second equation for V_o to get the initial speed?
    I got V_o=(h-0.5*g*t2^2)/t2
     
  15. Jan 21, 2014 #14

    berkeman

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    You should combine the 2 equations before solving for V_o. That way you should be able to group things to turn the t1 and t2 absolute times into a t=t1-t2 relative difference in time. What's an easy way to combine the 2 equations?
     
  16. Jan 21, 2014 #15
    Could you just plug in the first equation into the h of the second equation?
    -0.5gt1^2=-V_o*t2-0.5gt2^2
     
  17. Jan 21, 2014 #16

    berkeman

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    Yep, both equations were equal to -h, so you can set them equal to each other.
     
  18. Jan 21, 2014 #17
    Alright, so then V_o=(-0.5*g*T^2)/t2
     
  19. Jan 21, 2014 #18

    berkeman

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    Not quite. It's looking like there is no clean way to express this in terms of t=t1-t2 (or the T you used).

    Remember that (t1-t2)^2 is not equal to t1^2 - t2^2. Just multiply out the first one to see the extra terms.

    It may be that the answer just has to be expressed in terms of t1 and t2. Is there a way you can check the answer?
     
    Last edited: Jan 24, 2016
  20. Jan 21, 2014 #19
    No, my homework is online though, so I have as many chances to enter in the right answer as I need. I'm not really sure how it wants the answer formatted though.
     
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