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idku

  • one year ago

I got a question about vector problem. (Please, I just started don't through hard stuff at me).

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  1. idku
    • one year ago
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    Two balls are thrown from the same height as shown. The \(\color{red}{\rm red}\) ball is thrown with velocity \(\color{red}{ \vec v}\) in the vertical direction. The \(\color{blue}{\rm blue}\) ball is thrown with twice the speed of the red ball at an angle of 30\(^\circ\) with respect to the horizontal. Which ball will go higher? |dw:1440947270312:dw|

  2. idku
    • one year ago
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    The choices they gave me are: A) the red ball B) the blue ball C) the maximum height for both balls is the same D) it depends on the value of v but I really want to figure how to do these.

  3. Michele_Laino
    • one year ago
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    hint: the vertical speed of the red ball is V whereas the vertical speed of the blue ball is 2V*sin(30)=2*V/2=V

  4. idku
    • one year ago
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    V \(?\) 2V \(\sin(30)\) V \(?\) 2V \(\sin(30)\) V \(?\) 2V \((1/2)\) V \(=\) V \(\)

  5. idku
    • one year ago
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    So they will reach the same height, right?

  6. Michele_Laino
    • one year ago
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    yes!

  7. idku
    • one year ago
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    (regardless of the V)

  8. Michele_Laino
    • one year ago
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    yes!

  9. idku
    • one year ago
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    I mean V is still greater than zero, just from the problem. Not that we even need to know this. Thank you!! Just one more thing, so for a horizontal motion of some vector \(\bf v\), with a degree of \(\theta\), the ball will go a horizontal distance D of: \({\rm D}={\bf v}\times \sin(\theta)\)

  10. idku
    • one year ago
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    (right? )

  11. Michele_Laino
    • one year ago
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    in order to compute the horizontal distance, we have to consider the horizontal component of the speed of the blue ball, whose magnitude is: 2*V cos(30)=sqrt(3)*V

  12. Michele_Laino
    • one year ago
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    \[\Large {v_x} = 2V\cos 30 = 2V \cdot \frac{{\sqrt 3 }}{2} = \sqrt 3 V\]

  13. idku
    • one year ago
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    So, in general, if we had: |dw:1440947970856:dw| then the horizontal distance is: \({\rm D}={\bf v}\times \sin(\theta) \times {\bf v} \cos(\theta)\)

  14. Michele_Laino
    • one year ago
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    no, the horizontal distance is: \[\Large D = {v_x}\Delta t\] where \Delta t is the total flying time

  15. idku
    • one year ago
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    and Vx is what exactly?

  16. Michele_Laino
    • one year ago
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    Vx is: \[\Large {v_x} = V\cos \theta \]

  17. idku
    • one year ago
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    \(\large {\rm D_{horizontal }}=\left(\Delta t\right) {\bf v} \cos(\theta)\)

  18. Michele_Laino
    • one year ago
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    In general, referring to your last drawing, the total flying time is: \[\Large \Delta t = \frac{{2V\sin \theta }}{g}\]

  19. idku
    • one year ago
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    g is -9.8 m/s

  20. idku
    • one year ago
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    just clarifying for myself

  21. Michele_Laino
    • one year ago
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    so teh horizontal distance is: \[\Large D = {v_x}\Delta t = V\cos \theta \cdot \frac{{2V\sin \theta }}{g} = \frac{{{V^2}}}{g}\sin \left( {2\theta } \right)\]

  22. Michele_Laino
    • one year ago
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    the*

  23. idku
    • one year ago
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    yes, that makes sense:)

  24. Michele_Laino
    • one year ago
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    yes! g= 9.8 m/sec^2 or g= 32 feet/sec^2

  25. Michele_Laino
    • one year ago
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    :)

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