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butterflydreamer

  • one year ago

Projectile motion question. Please help me :). I'm stuck on part c. Initial velocity means at t=0, v=? Right? So if x = vt cos alpha, and vcos alpha= 4 Then x= 4t.. So wouldn't you just differentiate x=4t to find the velocity? So initial velocity = 4 ? Or is this a trick question?....

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  1. butterflydreamer
    • one year ago
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    Oh part c question 12 of the attachment :)

  2. Rhystic
    • one year ago
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    there is no attachment....

  3. butterflydreamer
    • one year ago
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    :/. I'll try add it again Can you see it now?

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  4. amistre64
    • one year ago
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    how do we prove part a? since a frisbee has different dynamics then say a ball or or a rock or an anvil.

  5. butterflydreamer
    • one year ago
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    hmm.. i used the same basic method as with a ball/rock/frisbee...

  6. amistre64
    • one year ago
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    a flying frisbee does not fall like a rock. the acceleration is altered by its motion of flight. ive thrown a frisbee, and a rock ... they dont have the same trajectories.

  7. amistre64
    • one year ago
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    in a vacuum cleaner they might ...

  8. amistre64
    • one year ago
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    the derivative of position will give us velocity, but im not sure what math you are able to use

  9. butterflydreamer
    • one year ago
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    I guess that's true but the question doesn't really mention that? Using the same general method, i still managed to prove part a: |dw:1433953611028:dw|

  10. butterflydreamer
    • one year ago
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    by the way, i wrote it wrong and my "theta" is meant to be an alpha LOL

  11. amistre64
    • one year ago
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    the x direction has no acceleration acting on it. so it is constant.

  12. butterflydreamer
    • one year ago
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    sorry but i don't quite understand what you mean... So am i wrong in saying v = 4 ?

  13. amistre64
    • one year ago
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    you are wrong for saying x' = something that varies with time. assuming a frictionless setup. you see, gravity pulls downward, not sideways. there is not force pushing this thing sideways. it is moving forward under its own inertia which was imparted to it at the start.

  14. butterflydreamer
    • one year ago
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    okay. Yes gravity is a downwards motion. I'm just confused on how else i would solve for velocity.

  15. butterflydreamer
    • one year ago
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    *initial velocity

  16. amistre64
    • one year ago
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    well, velocity id distance per time frame.

  17. amistre64
    • one year ago
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    the velocity is in reference to the distance covered, divided by the time it took to get there. if i travel 20 miles, in 1 hour ... my velocity is 20 mph

  18. amistre64
    • one year ago
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    the horizontal velocity is therefore: \[v_x=\frac{6}{1.5}\frac{m}{s}\]

  19. butterflydreamer
    • one year ago
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    which equals to 4m/s ...

  20. amistre64
    • one year ago
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    our velocity with respect to y, accelerates since it is affect by an up/down force ... the force of gravity

  21. amistre64
    • one year ago
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    yes, in part b we prove: v cos(a) = 4

  22. butterflydreamer
    • one year ago
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    i've already answered part a and b I just needed help with part c.

  23. amistre64
    • one year ago
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    what is 'exact form' ?

  24. phi
    • one year ago
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    the velocity can be represented as a vector <vx, vy> you want the magnitude of this vector

  25. amistre64
    • one year ago
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    you found the initial velocities in part b, what angle has an adjacent leg of 4, and an opposite leg of 15?

  26. phi
    • one year ago
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    I would do sqrt( v^2 cos^2 + v^2 sin^2) = sqrt(4^2 +15^2)

  27. amistre64
    • one year ago
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    the length of the vector, represents the intial velocity in the direction of the toss.

  28. butterflydreamer
    • one year ago
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    tan (a) = 15/4 (a) = 75 degrees (nearest degree) ?

  29. phi
    • one year ago
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    yes

  30. butterflydreamer
    • one year ago
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    alright, so we just substitute in (a) into Vcos (a) = 4 ? So V = 4/(cos 75) ?

  31. butterflydreamer
    • one year ago
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    so then velocity = 15.5 ?

  32. phi
    • one year ago
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    By "exact form" I think they want the sqr form

  33. butterflydreamer
    • one year ago
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    ohhh alright. Thank you!!! Sorry if i frustrated you guys , projectiles are confusing lol.

  34. phi
    • one year ago
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    btw, how do you show v sin a = 15 I don't see how

  35. butterflydreamer
    • one year ago
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    Well, max height occurs when dy/dt = 0 so since y = -5t^2 + vsin(a) t dy/dt = -10t + vsin(a) Then sub indy/dt = 0 , t = 1.5 0 = -10 (1.5) + vsin(a) vsin(a) = 15

  36. phi
    • one year ago
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    thanks!

  37. butterflydreamer
    • one year ago
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    no problem LOL

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