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anonymous

  • one year ago

A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 36t + 9. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball's maximum height? A) 1 s; 22 ft B) 2 s; 22 ft C) 2 s; 6 ft D) 1 s; 54 ft

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  1. arindameducationusc
    • one year ago
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    is this a physics question?

  2. Michele_Laino
    • one year ago
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    from your equation of motion, I see that the initial speed is 36 feet/sec

  3. Michele_Laino
    • one year ago
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    now, the equation for speed at a generic time t, is: \[\Large v\left( t \right) = {v_0} - gt \to v\left( t \right) = 36 - 32t\]

  4. Michele_Laino
    • one year ago
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    in order to compute the time t, at which the ball has reached its maximum height, you have to set v=0, so you have to solve this equation: \[\Large 0 = 36 - 32t\] please solve for t

  5. anonymous
    • one year ago
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    No, Algebra @arindameducationusc

  6. anonymous
    • one year ago
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    How am I supposed to do that.. can u help me please @Michele_Laino

  7. Michele_Laino
    • one year ago
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    the general formulas of kinematics, are: \[\Large \begin{gathered} h\left( t \right) = - \frac{1}{2}g{t^2} + {v_0}t + {h_0} \hfill \\ v\left( t \right) = {v_0} - gt \hfill \\ \end{gathered} \]

  8. Michele_Laino
    • one year ago
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    now, when the ball reaches its maximum height, its speed is zero, right?

  9. Michele_Laino
    • one year ago
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    from your equation of h(t), I can say that the initial speed is 36 feet/sec, and not 32 feet/sec

  10. anonymous
    • one year ago
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    Could there not also be a horizontal component of initial velocity such that the vertical component is 32 m/s yet and horizontal component is such that initial velocity is 36 m/s?

  11. anonymous
    • one year ago
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    From a mathematical perspective, a) is asking for the t-coordinate of the vertex. This is given by \[t=-\frac{ b }{ 2a }\]with b=36 and a=-16 from the given function.

  12. IrishBoy123
    • one year ago
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    none of the above https://gyazo.com/754cc35432c834269c8ad1f8315d4fbe

  13. IrishBoy123
    • one year ago
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    if this is algebra, you must be doing something like completing the square?? pls advise.

  14. anonymous
    • one year ago
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    quadratic equations and functions @IrishBoy123

  15. IrishBoy123
    • one year ago
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    so complete the square for \(h = –16t^2 + 36t + 9\)

  16. IrishBoy123
    • one year ago
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    helps to take the leading coefficient outside, as in \(\large h(t)=–16(t^2-\frac{36}{16}t-\frac{9}{16})\) lots of square numbers in there so should be fine...

  17. anonymous
    • one year ago
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    oh my

  18. anonymous
    • one year ago
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    my guess is that this has nothing to do with calculus it is really not this complicated \[ h = –16t^2 + 36t + 9 \] is a parabola that opens down the maximum is the second coordinate of the vertex the first coordinate of the vertex of \[y=ax^2+bx+c\] is always \[-\frac{b}{2a}\] which, in our case is \[-\frac{36}{2\times (-16)}=1\]

  19. anonymous
    • one year ago
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    ok not 1 sorry

  20. anonymous
    • one year ago
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    \[\frac{36}{32}=1.25\]

  21. anonymous
    • one year ago
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    that answers the first question a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.

  22. anonymous
    • one year ago
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    to find the maximum height replace \(t\) by \(1.25\) i see your answer choices do not have \(1.25\) so either there is a mistake in the answers, or it should have been \[h=-16t^2+32t+9\]

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