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  1. ksaimouli
    • 3 years ago
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    in that b and c

  2. ksaimouli
    • 3 years ago
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    @furnessj

  3. furnessj
    • 3 years ago
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    do you have answers to check?

  4. ksaimouli
    • 3 years ago
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    check bottom page

  5. ksaimouli
    • 3 years ago
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    i did a y=112 initial velocity

  6. ksaimouli
    • 3 years ago
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    y=vit-4.9t^2

  7. ksaimouli
    • 3 years ago
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    |dw:1352405626465:dw|

  8. ksaimouli
    • 3 years ago
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    is that right

  9. furnessj
    • 3 years ago
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    sorry i didn't look at the second page. First equation is simple: v = v_o +at 0 = 112 +(-g)t now, you need to change feet into metres to use 9.81 for g 112 / 3.28 = 34.146 so -34.146 / -9.81 = t

  10. furnessj
    • 3 years ago
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    Second part: you are only worried about horizontal velocity (no air resistance), the up and down plays no part here. So simply: horizontal distance = horizontal speed x time hd = 224cos30 x 7

  11. ksaimouli
    • 3 years ago
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    instead of v=v0 cani use v=v0t+1/2gt^2

  12. furnessj
    • 3 years ago
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    no, that equation is s = v0t + 1/2gt^2, and s is displacement, which we don't know yet

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