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anonymous

  • one year ago

help really stressed A juggler is performing her act using several balls. She throws the balls up at an initial height of 4 feet, with a speed of 15 feet per second. If the juggler doesn't catch one of the balls, about how long does it take the ball to hit the floor? Hint: Use H(t) = −16t2 + vt + s.

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  1. anonymous
    • one year ago
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    please help me

  2. Miracrown
    • one year ago
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    \[h(t) = -16t^2 + vt + S\] We have an initial height of 4 feet in the equation, this is s ..we are also given a speed of 15 feet/second

  3. Mehek14
    • one year ago
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    I think the quadratic function will work for this

  4. Mehek14
    • one year ago
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    after plugging in everything

  5. Miracrown
    • one year ago
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    in the equation, this is V and we are interested in the amount of time it iwll take for the ball to drop tot he ground at this time, height will be zero \[h(t)=−16t^2+vt+S \space = 0\]

  6. Miracrown
    • one year ago
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    so what we have here is a quadratic equation, we can solve this using the quadratic formula like @Mehek14 said

  7. Mehek14
    • one year ago
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    \(\bf{x=\dfrac{-b±\sqrt{b^2-4ac}}{2a}}\)

  8. Mehek14
    • one year ago
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    original formula is \(\bf{f(x)=ax^2+bx+c}\)

  9. Mehek14
    • one year ago
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    for a, you have -16 for b, you have 15 and for c, you have 4 @jaredhm29 can you plug that into the quadratic formula?

  10. Mehek14
    • one year ago
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    @jaredhm29 are you there?

  11. anonymous
    • one year ago
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    yes

  12. anonymous
    • one year ago
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    Ive had a stuggle in this segment

  13. anonymous
    • one year ago
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    @Mehek14

  14. Mehek14
    • one year ago
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    can you plug a, b , and c into the quadratic formula?

  15. anonymous
    • one year ago
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    idk what u mean sry

  16. Mehek14
    • one year ago
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    \(\bf{x=\dfrac{-b±\sqrt{b^2-4ac}}{2a}}\)

  17. Mehek14
    • one year ago
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    for a, you have -16 for b, you have 15 and for c, you have 4

  18. Mehek14
    • one year ago
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    \(\bf{x=\dfrac{-15±\sqrt{15^2-4*(-16)*4}}{2*(-16)}}\)

  19. Mehek14
    • one year ago
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    first thing what is \(15^2=?\)

  20. Mehek14
    • one year ago
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    @jaredhm29 ?

  21. anonymous
    • one year ago
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    @Theloshua

  22. anonymous
    • one year ago
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    please help me @Miracrown

  23. anonymous
    • one year ago
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    @zzr0ck3r @Miracrown

  24. anonymous
    • one year ago
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    @mathmate

  25. mathmate
    • one year ago
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    As @Miracrown said, \(\large h(t) = -16t^2 + vt + S\) is a standard kinematics equation giving the height h above datum (ground) at time t (therefore h(t)) as a function of t. v is the initial velocity (up = positive)=15 ft/s, and S= initial location (4 ft above datum= ground). -16 is actually acceleration due to gravity (32.2 ft/s^2) divided by 2, negative because acceleration is downwards. Substitute these values, you will get the same equation as @mehek14 showed you. Solving for t and reject the negative root gives you the time required.

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