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YumYum247

  • one year ago

Help me please!!! :D

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  1. anonymous
    • one year ago
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    \[H=\frac{u^2\sin^2(\theta)}{2g}\]

  2. YumYum247
    • one year ago
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    |dw:1441080373399:dw|

  3. anonymous
    • one year ago
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    Use the formula I wrote

  4. YumYum247
    • one year ago
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    but i din't know what u^2 means tho??????

  5. YumYum247
    • one year ago
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    initial velocity?

  6. anonymous
    • one year ago
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    yep

  7. YumYum247
    • one year ago
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    aight.....Thanks!!!

  8. YumYum247
    • one year ago
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    but that formula is used to find the maximum horizontal displacement tho!! i need the vertical displacement.

  9. YumYum247
    • one year ago
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    @welshfella please help me!!! :)

  10. anonymous
    • one year ago
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    That gives the maximum vertical displacement, the H stands for height, the formula is correct

  11. anonymous
    • one year ago
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    For horizontal displacement we have, \[R=ucos(\theta)T=ucos(\theta).(\frac{2usin(\theta)}{g})=\frac{u^2.(2\sin(\theta)\cos(\theta))}{g}=\frac{u^2\sin(2\theta)}{g}\]

  12. YumYum247
    • one year ago
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    sorry i didn't see the 2g at the bottom....hehehehehhehe :"D Thanks Luv!!

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