anonymous
  • anonymous
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?
Mathematics
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anonymous
  • anonymous
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?
Mathematics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
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
anonymous
  • anonymous
??
anonymous
  • anonymous
equation is \[h(t)=25t-16t^2\]

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anonymous
  • anonymous
maximum is given by \[\frac{-b}{2a}\] and here \[a=-16,b=25\] get \[\frac{-b}{2a}=\frac{25}{32}\]
anonymous
  • anonymous
there shouldnt be anything wrong with using v^2 -u^2 +2as
anonymous
  • anonymous
then plug it in
anonymous
  • anonymous
I hope my equation and calculation is correct.
anonymous
  • anonymous
maybe it is let me check
anonymous
  • anonymous
thank youuu guys :)
anonymous
  • anonymous
aadarsh the velocity is feet per second and you used m/s for gravity thats the only problem I think
anonymous
  • anonymous
m/s^2
anonymous
  • anonymous
i get \frac{1775}{64}=27.73 \]
anonymous
  • anonymous
\[\frac{1775}{64}=27.73\]
anonymous
  • anonymous
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
anonymous
  • anonymous
I did it both ways and got 625/64=9.77 feet both times
anonymous
  • anonymous
using 32 ft/s^2 for gravity

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