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idku

  • one year ago

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  1. idku
    • one year ago
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    ``` Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. ``` |dw:1441299455460:dw|

  2. idku
    • one year ago
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    theta = 44° and I need the distance between the two people

  3. geerky42
    • one year ago
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    Plan is to figure out the time it takes for ball to go from Juile to Sarah. For that, we focus on vertical component. After that, we can focus on horizontal component and figure out the distance.

  4. idku
    • one year ago
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    I have found so far: 1) Horizontal Component of the Velocity: \(\\[0.8em]\) \(V_x~~=~~V_0\times \cos(\theta)~~=~~11\times \cos(44)~~=~~11\cos(44)\) 2) Vertical Component of the Velocity: \(\\[0.8em]\) \(V_y~~=~~V_0\times \sin(\theta)~~=11\times \sin(44)~~=~~11\sin(44)\) 3) Maximum Height: \(\\[0.8em]\) \(V_y~~=~~V_{0y}-gt\) \(0~~=~~11\sin(44)-9.81t\) so max height ocucrs at \(t=11\sin(44)/9.81\) and now we plug that into vertical displacement: \(y=V_{0y}\times t+y_0\) \(y=11\sin(44)\times \left( 11\sin(44)~/~9.81\right)+1.5 \approx 4.5 ~~({\rm meters})\)

  5. idku
    • one year ago
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    What formula would I use to find the time that takes the ball to fly from one person to another?

  6. idku
    • one year ago
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    Disregard the previous long 3 part reply, that is more for myself.

  7. geerky42
    • one year ago
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    Because initial height and final height are same, we can just double the time it takes for ball to reach maximum height.

  8. IrishBoy123
    • one year ago
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    yeah, you don't need to know how high it went, just how long it was in the air for

  9. idku
    • one year ago
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    Can we go by the first thing you proposed please? \(\color{blue}{\text{Originally Posted by}}\) geerky42 Plan is to figure out the time it takes for ball to go from Juile to Sarah. For that, we focus on vertical component. After that, we can focus on horizontal component and figure out the distance. \(\color{blue}{\text{End of Quote}}\)

  10. idku
    • one year ago
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    Which formula am I starting to use, can you please tell me?

  11. geerky42
    • one year ago
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    You figured out the time it takes for ball to reach max, so just double it (2t), then we got time we need.

  12. idku
    • one year ago
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    ok, so we need `22sin(44) / 9.81` seconds, (roughly 4.5 × 2 = 9) seconds.

  13. idku
    • one year ago
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    ok, and then how would I use the time between them to find the distance between them>

  14. geerky42
    • one year ago
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    Just focus on horizontal component. We know that horizontal velocity doesn't change, so formula is simply \(d = vt\) we have time and horizontal velocity, so we can multiply them to get distance.

  15. geerky42
    • one year ago
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    Makes sense, right?

  16. idku
    • one year ago
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    oh, so d=11cos(44)×(11sin(44)÷9.81)= (11/19.62) sin(88)

  17. idku
    • one year ago
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    11cos(44) is v 11sin(44) / 9.81 is time

  18. geerky42
    • one year ago
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    actually time is 22sin(44)/9.81 remember we need time it takes for ball to go from Julie to Sarah.

  19. idku
    • one year ago
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    Ok, 12.32 is wolfram's approximation. How does it exactly work.....? d=vt Only gives us 1/2 of the distance for some reason?

  20. idku
    • one year ago
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    or rather full time is 2t.

  21. idku
    • one year ago
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    so t is 2t:)(()()()()

  22. geerky42
    • one year ago
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    Yeah

  23. idku
    • one year ago
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    Ok, I am very curious, because on my test it would probably not be this way. Can you expain how would you go about doing your original approach that you proposed? ``` Plan is to figure out the time it takes for ball to go from Juile to Sarah. For that, we focus on vertical component. After that, we can focus on horizontal component and figure out the distance. ```

  24. geerky42
    • one year ago
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    More of less same as what we just did. Horizontal component is very simple \(d = vt\), however we only know the value of \(v\). So I'd just calculate full time with given vertical velocity by using \(V_y = V_{o_y}−gt\) Initial velocity and final velocity should be opposite, right? So we have \[-V_{o_y} = V_{o_y}-gt\quad\Longrightarrow\quad\dfrac{2V_{o_y}}{g}=t\] Now we got \(t\), so we then calculate distance.

  25. idku
    • one year ago
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    Oh Vy=Voy - gt is what we would have used. Tnx

  26. geerky42
    • one year ago
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    No problem

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