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Question :A diver springs upward from a board Need help

  1. Sep 18, 2006 #1
    Question :A diver springs upward from a board ......Need help

    I am stumped as to how to go about answering this question...A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 8.90 m/s and her body makes an angle of 73.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction. I am not sure if I am to set this up as two different columns and solve for both the x and y components? Any help would be appreciated. thanks
     
  2. jcsd
  3. Sep 18, 2006 #2
    Absolutely solve for the x- and y-components separately.

    The only connection between them is that they will have the same elapsed time (which may or may not help you in this particular problem; I haven't looked that closely). And, of course, you need to combine the components to get the vector in question...
     
  4. Sep 18, 2006 #3
    I guess im just confused on how to set this problem up. The angle is throwing me off, am i solving for tan...or is it a sin cos im looking for? this whole problem has me confused.
     
  5. Sep 18, 2006 #4
    Have you drawn a diagram?

    Also, it sounds to me (correct me if I'm wrong) that you're stressing over the last step. Take things one step at a time. Make a list of what you know for x, what you know for y, what you know in general.
    List what you want to know (what the question asks for).
    Write down an equation that will give you what you want to know. If that equation requires you to sove for something else first, do so.

    I know, that's all up in the air (no[?] pun intended). Basically, have you drawn a diagram yet? Are you working in an organized fashion?
     
  6. Sep 18, 2006 #5
    Here's what i think is going on so far:
    y initial =0
    y final = -3
    im trying to draw a triangle to solve for the info im giving about her speed of 8.9 m/s and the angle with respect to the horizontal which is 73 degrees.
    But isnt her speed....gravity, -9.8 m/s 2??
    I have tried drawing this up....over and over
    It's just not connecting.
     
  7. Sep 18, 2006 #6
    OK, now we're in business.
    Let's start with resolving the diver's final velocity into its x and y components. If someone cool were helping you out here, they'd draw the diagram now. I don't know how to draw diagrams on the computer.
    Your diagram should be a parabola opening downward, with the bottom left bit chopped off (because she starts from an elevated diving board). The bottom right hand bit is where she enters the water. That forms a right triangle with an angle which is given as 73 degrees, and the hypotenuse is her final velocity, given as 8.9 m/s.
    The two legs of that triangle are the x and y components of her velocity. Use arccos and arcsin to solve for the components.
    Does that help?

    Nope. Gravity is her acceleration.
     
  8. Sep 18, 2006 #7
    I tried finding the arccos and arcsin....but it doesnt seem right.
    I got arcsin= 9.3 and arccos = 30.4.??
    And after i find those (when i get them right), ill have
    y initial= 0
    y final = -3
    velocity component of y = (whatever the correct answer from above is)
    velocity component of x= ( " " ")
    acceleration = -9.8 m/s2
    Did i miss anything?
     
  9. Sep 18, 2006 #8
    yeah, it shouldn't seem right; there's no reason to start playing with arccos and arcsin over here.

    You have a right triangle with angle theta = 73 degrees.
    [itex]cos\theta = \frac{adjacent}{hypotenuse} = \frac{v_{x}}{v}[/itex]
    so [itex]v_{x} = v\cos\theta[/itex]
    so in this case [itex]v_{x} = (8.90 \frac{m}{s})(cos\73^o)[/itex]

    Then do a similar thing to find the y-component.
     
  10. Sep 18, 2006 #9
    ok....so i come up with ...
    Vox= 2.6 m/s
    Voy= 8.5 m/s
     
  11. Sep 18, 2006 #10
    You're looking for initial velocity. Find the x and y components seprately. What equations could you use?
     
  12. Sep 18, 2006 #11
    So those answers i came up with were wrong?
    Im sorry.....I thought that was what i was to find, in the previous step?
    Im sorry, i am confusing myself big time
     
  13. Sep 18, 2006 #12
    Whoa....
    No, I'm the one who's not being clear; sorry. What you just found are the correct components of the FINAL velocity. Cute. But the question wants initial velocity. So you need to find an equation relating initial and final velocities (preferably without time, because you don't know it yet) and use it to find the initial velocity. start with finding the y-component of the INITIAL velocity.
    Then I'd use that to find time, and use time to find the x-component of the initial velocity.
     
  14. Sep 18, 2006 #13
    So then ill try using V*2= Vo*2 + 2a(X-Xo)
    to solve for initial velocity.
     
  15. Sep 18, 2006 #14
    but for the Y...not the x component....sorry about that
     
  16. Sep 18, 2006 #15
  17. Sep 18, 2006 #16
    well i came up with a y-component of the initial velocity= 3.66 m/s
    then i used Vy=Voy + at to find time
    but i came up with -.49....???
    so i tried plugging that in to find the initial velocity of x-component.....using V= Vo +at.....but i got 2.6???
    That isnt right....:(
     
  18. Sep 18, 2006 #17
    where am i going wrong??....thakn you for your patience by the way...much appreciated.
     
  19. Sep 18, 2006 #18
    Well, getting negative time is (usually) a pretty good indication you did something wrong. I got a different initial y-velocity than you did. How did you manipulate your equation?

    (and my pleasure with the help. really. this is so much fun! [quoth the nerd])
     
  20. Sep 18, 2006 #19
    well like i said, i started with Vy*2= Voy*2 + 2a (Y-Yo)
    so i plugged in numbers...(8.5)*2= Voy*2 + 2(-9.8)(-3-0)
    and i got 13.45=Voy*2
    for an answer of Voy= 3.66
     
  21. Sep 18, 2006 #20
    OK, I manipulated the equation wrong. Divided instead of subtracted. BUT. I then did it again your way (read "the right way"), using v_y=-8.51 m/s. (NOTE that the y component of the final velocity is negative. It points down, and you've set your co-ordinate system so down is negative.)This gives v_0y as 3.68 m/s.
    Great. Now shove that into, say, [itex]v_{y}=v_{0y}+a_{y}t[/itex]
    and get 1.24s. Please check my work, though, it's late where I live...
    Use that time instead of the negative one, and see if that helps.
     
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