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Homework Help: Projectile motion question and minimum initial speed

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Assume acceleration due to gravity is 9.8ms^-2

    It has been said that in his youth George Washington threw a silver dollar
    across a river. Assuming that the river was 75m wide,

    (a) what minimum initial speed was necessary to get the coin across the river
    (b) how long was the coin in flight?
    Ans (a) 27.1 m/s (b) 3.91 s

    2. The attempt at a solution

    i dun even know where to start with 2 variables =(
    only tried assuming his height =o
     
  2. jcsd
  3. Jan 19, 2010 #2

    Borg

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    This is a twist on your question from yesterday. To start this one, ask yourself what the optimal angle would be to throw a coin (assuming no air resistance). Then look at the formulas that you know to see if you can determine the flight time.
     
  4. Jan 19, 2010 #3
    =O u remembered
    thx for the help again =D

    Assuming angle to be 45 degrees,
    maximum height i tink is => tan45 = h / 75
    h = 75m
    s = ut + 1/2at^2
    t = sqrt(75/4.9)
    = 3.912303982
    = 3.91s which answered part b

    but is there anyway to get part a 1st before part b =O
    mayb there is a way now with the angle =o

    is this wrong?
    i use v^2 = u^2 + 2as for the vertical velocity
    v = 0, when it hit the water,
    u^2 = 2(9.8)(75)
    u = sqrt(ANS)
    = 38.34057903m/s

    then i tought that if 45 degrees, Vx = Vy, initial velocity u = sqrt(2Vy^2)
    u = sqrt(2*38^2)
    u = 54.22176685 which is twice of the answer =o
     
    Last edited: Jan 19, 2010
  5. Jan 19, 2010 #4

    Borg

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    Hi again. You got the right answer but, you mislabeled one item. The maximum height is not 75m - that is the distance across the river. The equation for the X-direction is:

    s = ut where u is the velocity in the x direction

    In the s = ut + 1/2at^2 equation, u is the vertical direction. But, since both velocities start with the same value, you know that ut = 75 from the first equation.

    As far as solving it to get the time last, I think they do that to throw you off.
     
  6. Jan 19, 2010 #5
    =o ic

    but is it possible to assume the maximum height to be at half of the river's width? =o
     
  7. Jan 19, 2010 #6

    Borg

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    Yes. Can you tell me what it would be?
     
  8. Jan 19, 2010 #7
    i guess 37.5m =o both the height and half-width
     
  9. Jan 19, 2010 #8

    Borg

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    Don't guess. :smile: What is the vertical velocity at it's highest point?
     
  10. Jan 19, 2010 #9

    Borg

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    I think we're confusing each other. I'm asking what the maximum height is and not where the max occurs.
     
  11. Jan 19, 2010 #10
    lol now im stuck again is it zero?

    i cant do it =o i only used 37.5 * tan45 lol = 37.5 oo
    how do u find maximum height lol
     
  12. Jan 19, 2010 #11

    Borg

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    What goes up, has to come down. Remember your first two equations. One applies to the horizontal direction and the other to the vertical. [STRIKE]What is s when u = 0?[/STRIKE]
     
    Last edited: Jan 19, 2010
  13. Jan 19, 2010 #12
    first equation => s = ut (horizontal)
    second equation => s = ut + 1/2at^2 ( vertical )
    s = 1/2at^2? =o
    but i still need to find t 1st =o

    man im confused now i dun even know how to find t =o
     
  14. Jan 19, 2010 #13

    Borg

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    I was afraid that I would confuse you with that last question. Sorry, let's take a step back. Think about your problem from yesterday. Yesterday, you knew the height and wanted to solve for time. Now you have the time (remember - halfway across) and you want to know the height. Remember that yesterday's initial vertical velocity was zero?
     
  15. Jan 19, 2010 #14
    lol im sorry for taking up your time =o

    but with the equation s = 1/2at^2, as u = 0.
    s = 1/2(9.8)(3.91)^2 which = 74.91169 ~ 75m o_O

    im still guessing the t should be divided by 2
    den max h = 75/2 = 37.5 =O
    =o nvm u mentioned halfway across =D

    is it correct =D
     
    Last edited: Jan 19, 2010
  16. Jan 19, 2010 #15

    Borg

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    I'm new at this as well but, hang in there and we'll get finished.

    Take another look at s = ut + 1/2at^2. What are these things? s is the height above the ground. You solved the previous equation because you knew that when the coin returned to earth, s = 0. Now you want to solve for s at the halfway point (t = 3.91s/2). From what you have solved already, you know that the initial velocity is 27.1. Therefore, you have the following:

    s = 27.1t + (1/2)(-9.8)t^2

    Plug in t for the halfway point.
     
  17. Jan 19, 2010 #16
    its ok ill try to clear my mind and do it at a later time thank you so much for all your help =D
     
    Last edited: Jan 19, 2010
  18. Jan 19, 2010 #17

    Borg

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    I think that I really confused you with the zero velocity comment. The maximum height is 34.25 as you've calculated.
     
  19. Jan 19, 2010 #18
    o ok now i just need to redo part a and b cant find height without these 2 anyway lol

    im off to bed cya =D and i really appreciate you taking time out to help me =D sometimes too much just isnt enough coz i really need time to digest all this
     
    Last edited: Jan 19, 2010
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