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anonymous

  • 5 years ago

Jules kicks a soccer ball off the ground and into the air with an initial velocity of 25 feet per second. Assume the starting height of the ball is 0 feet. Approximately what maximum height does the soccer ball reach?

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  1. anonymous
    • 5 years ago
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    By Newton's third law of motion, v^2=u^2+2as where, "v" denotes final velocity, "u" denotes initial velocity, "a" denotes acceleration, and "s" denotes the height attained. Here, "v"=0. Therefore, (-u)^2=2as Or, -25*-25=2*10*s Or, 625=20s Or, s=625/20=31.25 feet

  2. anonymous
    • 5 years ago
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    ??

  3. anonymous
    • 5 years ago
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    equation is \[h(t)=25t-16t^2\]

  4. anonymous
    • 5 years ago
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    maximum is given by \[\frac{-b}{2a}\] and here \[a=-16,b=25\] get \[\frac{-b}{2a}=\frac{25}{32}\]

  5. anonymous
    • 5 years ago
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    there shouldnt be anything wrong with using v^2 -u^2 +2as

  6. anonymous
    • 5 years ago
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    then plug it in

  7. anonymous
    • 5 years ago
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    I hope my equation and calculation is correct.

  8. anonymous
    • 5 years ago
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    maybe it is let me check

  9. anonymous
    • 5 years ago
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    thank youuu guys :)

  10. anonymous
    • 5 years ago
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    aadarsh the velocity is feet per second and you used m/s for gravity thats the only problem I think

  11. anonymous
    • 5 years ago
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    m/s^2

  12. anonymous
    • 5 years ago
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    i get \frac{1775}{64}=27.73 \]

  13. anonymous
    • 5 years ago
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    \[\frac{1775}{64}=27.73\]

  14. anonymous
    • 5 years ago
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    ya, ya!! ur correct. Let me do the correction. Or, -25*-25=2*10*10/3*s Or, 625=200/3s( Since, a=10 m/s^2, now, 1m=100 cm, and 30 cm=1 foot, so, 1 m=10/3 feet) Or, 625*3=200s Or, 1875=200s Or, s=1875/200=9.375 feet

  15. anonymous
    • 5 years ago
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    I did it both ways and got 625/64=9.77 feet both times

  16. anonymous
    • 5 years ago
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    using 32 ft/s^2 for gravity

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