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Need help on direction of acceleration

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data
    The bridge is 9.5 m above the water and using his watch, Andrew measured the time it took for a stone, thrown by David to return to the height of the bridge to be .5 seconds.

    a. what is the vertical velocity with which the stone left David's hand?
    b. how high did the stone go over the water?
    c. what was the magnutude of the stones velocity in the horizontal direction if it left david's hand at an angle of 38 degrees?
    d. what distance from the base of the bridge did the stone travel?
    2. Relevant equations
    s=ut+1/2at^2



    3. The attempt at a solution
    The thing I am confused about is whether to put the acceleration negative or positive.
    I let Vs=0
    and then
    0=ut+1/2at^2
    0=0.5u-9.8*1/2*(0.5)^2
    u=2.45 m s^-1

    for part b) I did
    At max height v=0
    0=2.45^2+2*-9.8*s
    s=0.306
    so max height is 9.5+0.306?= 9.806 m

    c)
    tan 38=2.45/uH
    uH=3.1358 ms^-1

    d) sH=uH*t
    3.1358*t
    v^2=2.45^2-2*9.8*-9.5
    ....
    and then I get a negative answer that cannot be square rooted :\

    Is what I'm doing right? I'm not sure on the direction of acceleration :\
     
    Last edited: Feb 25, 2012
  2. jcsd
  3. Feb 26, 2012 #2

    I like Serena

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    Hi! :smile:

    Shouldn't you just fill in t=0.5 sec here?

    What are you calculating here?
     
  4. Feb 26, 2012 #3
    Hi :)
    I thought t=0.5 was the time it takes for the ball to go up, and then reach the same height as the bridge. So with the Sh=Uh*t formula shouldn't the "t" be total time it takes for the ball to fall down again?

    On that thing, I am trying to find the final velocity so I can find the total "t"
     
  5. Feb 26, 2012 #4

    I like Serena

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    Well, you're problem statement isn't very clear, but I was thinking that "thrown by David to return to the height of the bridge" means that David threw the stone from the height of the bridge, and that the stone lands at the height of the bridge.
    At least that is how you calculated (a), (b), and (c) if I'm not mistaken.
    In particular it would mean that 0.5 seconds is the total time to go up and go down again.

    Or am I misunderstanding the problem statement?
     
  6. Feb 26, 2012 #5
    I'm a bit confused because the ball needs to go down to the water so it will go up, then go back to the same height as the bridge, and then go back down 9.5 m to the water...so total time shouldn't be 0.5 seconds
     
  7. Feb 26, 2012 #6

    I like Serena

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    Ah, okay.
    Btw, is it a stone or a ball? ;)

    So the ball would go up for 0.25 seconds, down again for 0.25 seconds at which time it passes the bridge, and then further down until the water, yes?

    So what you would like to know is the total time, so you can use that in your formula sH=uH*t, right?

    How much time does it take the ball to drop from its highest point to the water?
     
  8. Feb 26, 2012 #7
    Oh! so I should find the time to drop from the highest point and then add 0.25?
     
  9. Feb 26, 2012 #8

    I like Serena

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    Yep. That's the easiest method I can think of.

    http://www.circuitgallery.com/media/catalog/product/cache/1/small_image/9df78eab33525d08d6e5fb8d27136e95/M/a/MacMahon_abc005_250.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
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