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Study23 Group Title

Please, help? An Olympic diver springs off a high dive that is 3m above the surface of the water. When she lands in the water she is traveling at a speed of 8.9 m/s at an angle of 75.0° with respect to the horizontal. What was her take off speed and direction?

  • one year ago
  • one year ago

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  1. Study23 Group Title
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    @hartnn ?

    • one year ago
  2. Study23 Group Title
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    @AllTehMaffs

    • one year ago
  3. AllTehMaffs Group Title
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    picture!

    • one year ago
  4. AllTehMaffs Group Title
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    |dw:1382313081201:dw|

    • one year ago
  5. AllTehMaffs Group Title
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    First off, what do you know about her velocity in the x direction for the entire trajectory?

    • one year ago
  6. Study23 Group Title
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    It's constant?

    • one year ago
  7. AllTehMaffs Group Title
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    yup yup. So now all you need to find is her y velocity at 3m above the water, and you can figure out the direction and magnitude of initial takeoff , yah? How might one find that?

    • one year ago
  8. Study23 Group Title
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    Ummm... I'm not sure

    • one year ago
  9. AllTehMaffs Group Title
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    You have kinematics or energistics as your two options. Pick your poison.

    • one year ago
  10. AllTehMaffs Group Title
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    or both

    • one year ago
  11. AllTehMaffs Group Title
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    just not neither

    • one year ago
  12. Study23 Group Title
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    Kinematics is what we're learning

    • one year ago
  13. AllTehMaffs Group Title
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    that's probably the better option then :) What do you know about the diver's y velocity at the top of the trajectory?

    • one year ago
  14. Study23 Group Title
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    8.9sin75° ?

    • one year ago
  15. AllTehMaffs Group Title
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    That's right at the bottom, when she hits the water. What about right here?|dw:1382313886798:dw|

    • one year ago
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    Ohh! 0

    • one year ago
  17. AllTehMaffs Group Title
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    !!! :D

    • one year ago
  18. AllTehMaffs Group Title
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    exactly

    • one year ago
  19. AllTehMaffs Group Title
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    Did it all fall into place and the big picture come into focus? Or do wanna keep working it together?

    • one year ago
  20. Study23 Group Title
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    I'm still lost tbh

    • one year ago
  21. AllTehMaffs Group Title
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    none worries. Can you come up with an equation of motion that'll give you this height?|dw:1382314187399:dw| One that only uses velocity, distance, and acceleration?

    • one year ago
  22. Study23 Group Title
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    I'm not sure without t

    • one year ago
  23. AllTehMaffs Group Title
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    There's 1 of em is all. It's \[ v^2 = v_o^2 + 2ad\] v = final velocity vo = initial velocity a = acc. d = displacement

    • one year ago
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    Hmmm I haven't even seen that yet!

    • one year ago
  25. AllTehMaffs Group Title
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    That's the one I always forget, and always forget how to derive, too :P You get it (I'm neglecting a few terms, sorry) from \[ d = \left( \frac{v_o + v}{2} \right)t \] \[ t = \left( \frac{2}{v_o + v} \right)d \] then plug that into \[v = v_o + at\] It's mighty handy. Anyways... Once you get height H, what can you do?

    • one year ago
  26. Study23 Group Title
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    I could use this equation to find the velocity? But how would I find h?

    • one year ago
  27. AllTehMaffs Group Title
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    you know the velocity though! \[v = v_y= 8.9 sin \theta\] at the bottom, and vo = 0 at the top.

    • one year ago
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    Oh okay.

    • one year ago
  29. Study23 Group Title
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    So below the diving board (or would it be below the surface of the water) the \(\ \large v_y = 8.9sin\theta \text{ ? }\)

    • one year ago
  30. AllTehMaffs Group Title
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    exactly at the surface of the water Vy = 8.9sin75. We're trying to see, starting from rest, how far would the diver have to fall to be going that fast.

    • one year ago
  31. AllTehMaffs Group Title
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    Does that make sense? I tend to make things more convoluted than they need to be :/

    • one year ago
  32. Study23 Group Title
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    Yes, I think so. Where would I go from here?

    • one year ago
  33. AllTehMaffs Group Title
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    with that, we have the last bit of info. |dw:1382315519387:dw| When we have H, we us that same equation to find out the magnitude of the velocity height "little h," which is in line with the diving board. Then using that value you can find the unknown initial angle φ (phi) using tanφ = y/x . |dw:1382315666658:dw| The magnitude of the initial velocity is then \[ |\textbf v| = \sqrt{x^2+y^2}\] Follow?

    • one year ago
  34. AllTehMaffs Group Title
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    The answer takes the form, "Initial velocity was ...m/s at .... degrees above the horizontal."

    • one year ago
  35. Study23 Group Title
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    Okay, just one moment...Im trying to understand this all...!

    • one year ago
  36. AllTehMaffs Group Title
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    ~~ The first paragraph should say "... find out the velocity ~~ at~~ height 'little h'..."

    • one year ago
  37. Study23 Group Title
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    Okay that part of the velocity being same in direction but opposite in magnitude seems to make sense.... I'm a bit confused as to how to find x and y, though

    • one year ago
  38. AllTehMaffs Group Title
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    you have x already. It's stayed the same throughout the trajectory. You get y by using that 4th equation of motion again. \[ v_{y@h}^2 = 2ah\] Then \[\tan \phi = \frac{y}{x} = \frac{v_{y@h}}{8.9\cos7.5}. \]

    • one year ago
  39. AllTehMaffs Group Title
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    oh, y and x components of the triangle. Not x and y as in displacement. Sorry :P

    • one year ago
  40. AllTehMaffs Group Title
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    meant Vy and Vx

    • one year ago
  41. AllTehMaffs Group Title
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    `~~ \[ tan \phi = \frac{v_{y@h}}{8.9 \cos75} \]

    • one year ago
  42. AllTehMaffs Group Title
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    Hey, I gotta run. Sorry. I'll be on later :) good luck and good physicsing! Sciencing with friends is fun.

    • one year ago
  43. Study23 Group Title
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    Ok! Thanks for the help so far! :)

    • one year ago
  44. AllTehMaffs Group Title
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    change of plan. back. how's it goin?

    • one year ago
  45. Study23 Group Title
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    I'm really stuck actually @AllTehMaffs!

    • one year ago
  46. Study23 Group Title
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    I'm not sure which triangle x and y correspond to!

    • one year ago
  47. AllTehMaffs Group Title
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    My bad, I kept throwing around variables willy nilly. |dw:1382318816931:dw| the triangle x and y - which I should have said Vx and Vy - correspond to is the triangle where V is the hypotenuse. they're the x and y components of the initial velocity. \[ v_ix = v_i \cos \phi \] \[ v_iy = v_i \sin \phi \] We just happen to know them already, and are calculating Vi based on them. We know the initial velocity in the x direction is the same as the final x velocity down by the water \[ v_ix = 8.9\sin75 \] and the y velocity is what we found at little h - the velocity 3 meters above the water. \[ v_iy = v_{y@h} = \sqrt{2ah}\]

    • one year ago
  48. Study23 Group Title
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    This is starting to make some sense! What I'm confused about is how do I know what a and h are? Would h just be -9.8? Wait no 9.8? As for h... I'm not sure.

    • one year ago
  49. AllTehMaffs Group Title
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    h = H - 3 meters |dw:1382320100943:dw| , and a is -9.81m/s^2. I'm playing kind of loose with the signs ;P . When we find H: \[ v_f^2 = 2aH\] we're trying to see how big H is. We know H is a positive number, so we fix the units to be positive. This is a sloppy convention on my part. In that equation what we're really saying is |dw:1382319969777:dw| \[ v_f^2 = 2(-9.81)(-H)\] That's how the units technically work. Same for little h |dw:1382320225339:dw| We're measuring a distance below where we start "measuring." ...

    • one year ago
  50. Study23 Group Title
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    Ahhh Okay

    • one year ago
  51. AllTehMaffs Group Title
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    :)

    • one year ago
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    So y is equal to \(\ \Large \sqrt{2 \times -9.8 \times h \text{ ?}} \) \[\ \Large \text{What would h be then? I know it's H-3, but what's H? } \]

    • one year ago
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    @AllTehMaffs ?

    • one year ago
  54. Study23 Group Title
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    Wow, Thank You so much for all your help @AllTehMaffs! I finally understood how to find h, by first finding H using v^2f=2aH. I found h to be 1.037 (H=4.037), and I finally got an answer, which I think makes sense!!!! I got initial velocity was 5.06 m/s at 62.9 degrees above the horizontal. Thanks again @AllTehMaffs! :) :)

    • one year ago
  55. AllTehMaffs Group Title
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    @Study23 Sooo close! All of the concepts are there, just make sure that you realize in the equation \[ v_f^2=2aH\] (which, again, I wrote while using variables willy nilly) **the only velocity that contributes to H, the maximum height the diver gets to, is the Y velocity!** so it should be \[v_{f}^2 = v_o^2 + 2ad \] \[v_{fy}^2 = \cancel{v_{oy}^2} + 2a(H) \] \[v_{fy}^2 = 2a(H) \] So H is \[ H =\frac{v_{fy}^2}{2a}=\frac{{(8.9m/s\sin75)}^2}{(2)(-9.81m/s^2)}=-3.77m\] This equation says, "Starting from rest, a body must travel -3.77m - it must fall 3.77m meters below an initial location y=0 to attain a velocity of (8.9 sin 75)m/s given that acceleration is -9.81 m/s^2. A very wordy confusing say of looking at it!... which we can get around by saying everything is a magnitude. Then the equation reads as "Starting from rest, a body under acceleration 9.81 m/s^2 has a velocity of (8.9 sin 75)m/s in the direction of acceleration after it has displaced 3.77m in the direction of acceleration" Then all of the positive and negative information is retrieved from defining your coordinate system on the picture. . We can now say that the max height above the diving board , little h, is \[ h = H - 3 = .77m\] With that we're looking for the velocity of the diver after displacing .77 m \[ v_{y@h}=\sqrt{2ah} = \sqrt{(2)(9.81m/s^2)(.77m)} = 3.76 m/s\] "Starting from rest, after a body being accelerated at 9.81m/s^2 displaces .77m in the direction of the acceleration, it will be traveling a velocity of 3.76m/s in that direction." Therefor, starting at the top of the trajectory and falling .77m (going .77m in the -y direction), its velocity will be 3.76 m/s in the negative y direction. THIS means that at the other side of the parabola, its velocity will be equal in magnitude and opposite in direction! It's initial velocity in the y direction is 3.76 m/s in the positive y direction. Now we have v_xi and v_iy \[ v_{xi} = v_{xf} = 8.9 \cos 75 m/s\] \[v_{yi} = v_{y@h} = 3.76 m/s\] Using these values to find our angle phi \[ tan \phi = \frac{v_{yi}}{v_{xi}} = \frac{3.76m/s}{8.9 \cos 75 m/s}=1.59 \] \[ \phi = arctan 1.56 = 57.34º\] Now to find the magnitude of v_i, which is the square root of the sum of the squares of its components. \[ |\textbf v_i| = \sqrt{|v_{xi}|^2+|v_{yi}|^2} = \sqrt {(8.9\cos75m/s)^2+(3.76 m/s)^2} = 4.4m/s \] SO! ::breaths:: v_i = 4.4m/s @ 57.34º above the horizontal. This makes physical sense. The angle entering the water should be steeper than the initial angle because of an increase in the magnitude of the v velocity due to gravity.

    • one year ago
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