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Compassionate

  • one year ago

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t2 + 40ft + 1.5. Find the maximum height attained by the ball.

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  1. Mehek14
    • one year ago
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    \(\tt{x=\dfrac{-b}{2a}}\)

  2. Compassionate
    • one year ago
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    \[\frac{ -b }{ 2a}\] \[\frac{ -40 }{ -32 }\] Time = 8

  3. SolomonZelman
    • one year ago
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    What is your equation exactly?

  4. SolomonZelman
    • one year ago
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    \(h(t)= -16t^2 + 40t + 1.5\) ?

  5. SolomonZelman
    • one year ago
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    the max height will be the vertex (because the parabola opens down, and that is because the leading coefficient is negative)

  6. Compassionate
    • one year ago
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    I need to find the maximum height. Yes, that's the equation. So my time is 8, right? To find my height, plug 8 in.

  7. SolomonZelman
    • one year ago
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    You don't need the height, if you are given that function you can find the vertex and that will be your maximum height.

  8. Compassionate
    • one year ago
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    To find max height, I need to first obtain the time it takes the ball to reach it, then plug in t.

  9. Mehek14
    • one year ago
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    how did you get time = 8? \(\tt{\dfrac{-40}{-32}=1.25}\) not 8

  10. Compassionate
    • one year ago
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    I got 8 because my calculator is autistic and I am jut realizing this. -16(1.25)^2 + 40(1.25) + 1.5 -25 + 50 + 1.5 25 + 1.5 Height = 26.5

  11. Mehek14
    • one year ago
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    maximum height is 26.5 you get the time 1.25 and plug it in for t

  12. Compassionate
    • one year ago
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    Thanks! I got it :) `

  13. Mehek14
    • one year ago
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    when you use \(\tt{x=\dfrac{-b}{2a}}\), you find the time then you plug the time in for the variable and yw :]

  14. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle h(t)=-16t^2-40t+1.5 }\) \(\large\color{black}{ \displaystyle h(t)=-16(t^2+2.5t)+1.5 }\) side caclulation: \(\large\color{black}{ \displaystyle 2.5\div2=1.25 }\) \(\large\color{black}{ \displaystyle 1.25^2=1.5625 }\) \(\large\color{black}{ \displaystyle h(t)=-16(t^2+2.5t+1.5625-1.5625)+1.5 }\) \(\large\color{black}{ \displaystyle h(t)=-16(t^2+2.5t+1.5625)+(-16)(-1.5625)+1.5 }\) \(\large\color{black}{ \displaystyle h(t)=-16(t+1.25)^2+25+1.5 }\) \(\large\color{black}{ \displaystyle h(t)=-16(t+1.25)^2+26.5 }\) So the height is 26.5 and it occurs at t=1.25.

  15. SolomonZelman
    • one year ago
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    (via completing the square)

  16. Compassionate
    • one year ago
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    I never learned how to complete squares :(

  17. Compassionate
    • one year ago
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    Okay, so can I shoot one more at you guys?

  18. Mehek14
    • one year ago
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    don't you learn completing the square in Alg 1?

  19. SolomonZelman
    • one year ago
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    I don't think there is a single particular order for learning different concepts in math.... I have seen people who asked how to find a slope in DE section.

  20. SolomonZelman
    • one year ago
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    Well, completing the square is something that you can probably browse and find a good resource quite fast:)

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