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idku

  • one year ago

Please help me with 2 brief questions.

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  1. idku
    • one year ago
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    A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s. ------------------------------------------------------------ \(\rm \large I~found~so~far:\) [1] initial speed, \(\rm u=41.4(m/s)\) [2] angle of cannonball with respect to ground, \(\rm \theta=37.15º\) [3] maximum height the cannonball goes above the ground \(\rm x=31.8(m)\) [4] horizontal displacement is \(\rm 167.6(m)\) [5] the entire time of the ball motion \(\rm 5.1(s)\) ------------------------------------------------- \(\rm \large I~need:\) [6] speed of the cannonball at, \(\rm t=3.7(s)\) [7] height (Vertical Displacement) at, \(\rm t=3.7(s)\)

  2. anonymous
    • one year ago
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    6.Vi = √[Vx²+Vy²] = 44.2 m/s 7.3) Hmax = Vi²sin²Θ/(2g) = 62.49 m

  3. idku
    • one year ago
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    Are you sure? Becsause that doesn't seem to be right.

  4. anonymous
    • one year ago
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    Don't know,I never liked physics. Maybe he can help @Hero @dan815 @iambatman

  5. anonymous
    • one year ago
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    @triciaal

  6. idku
    • one year ago
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    I found the vertical displacement x=25.35m applying: \(x = ut + \dfrac{1}{2}at^2\) I have put: t=3.7s a=-9.81m/s\(^2\) u=25m/s

  7. anonymous
    • one year ago
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    Nice.

  8. idku
    • one year ago
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    I need velocity at t=3.7 the rest i found. tnx for coming btw

  9. anonymous
    • one year ago
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    Your welcome!

  10. dan815
    • one year ago
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    A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s. [6] speed of the cannonball at, t=3.7(s) [7] height (Vertical Displacement) at, t=3.7(s)

  11. idku
    • one year ago
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    [7]=25.35m found that:)

  12. dan815
    • one year ago
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    |dw:1441344582144:dw|

  13. dan815
    • one year ago
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    speed is the magnitude of velocity

  14. dan815
    • one year ago
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    velocity tells you how speed is broken into components of direction

  15. idku
    • one year ago
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    \(\large V_y=25-9.81(3.7)=-11.297\) \(\large V_x=33\) (yes, it doesn't change b/c acceleration is 0) \(\large V_t=\sqrt{33^2+11.297^2}=34.88\)

  16. dan815
    • one year ago
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    yeah

  17. dan815
    • one year ago
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    |dw:1441344791089:dw|

  18. idku
    • one year ago
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    oh, that is what I wanted to do. to use pythagorean theoream after fiding vert and horiz velocities (well horiz veloc I knew...:D) I was getting NEGATIVE 11.297 so I was thinking it is wrong.

  19. idku
    • one year ago
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    I mean those veloc at t=3.7

  20. dan815
    • one year ago
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    negative velocity just means down, it an indication of direction thats all

  21. idku
    • one year ago
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    I will take a note of that in my brain. Ok, so negative velocity is also an acceptable result (unlike displacement)

  22. dan815
    • one year ago
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    to someone who flipped the page, he will think the Veocity is positive it doesnt matter to him its the way u chose your coordinates

  23. idku
    • one year ago
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    I want to flip the page, but another page. The page of the physics intersecting w/ my life.... I don';t even know what I would do without you

  24. dan815
    • one year ago
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    LOL

  25. dan815
    • one year ago
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    get used to it, uve intersected with physics from the day u were born

  26. idku
    • one year ago
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    yes, but not with due dates

  27. idku
    • one year ago
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    and not with rediculous teachers such as my prof at college

  28. dan815
    • one year ago
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    xD

  29. idku
    • one year ago
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    In any case, I guess - Alright. I have a report on a lab. That I should be able to do. (Got 6 hours to work on it before my next classes)

  30. idku
    • one year ago
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    Have a good night or where ever yo are. Niagara falls:D

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