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anonymous

  • one year ago

An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t2 + 20t + 65. What is the object's maximum height?

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  1. Zale101
    • one year ago
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    You can find the maximum height of the object from the vertex. Do you remember how to find the vertex of an equation? |dw:1434946064036:dw|

  2. anonymous
    • one year ago
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    i really don't know i'm sorry i'm so bad at math

  3. Zale101
    • one year ago
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    You can either complete the square or find the vertex by this formula \(\Large Vertex=(h,k)=(\LARGE-\frac{b}{2a},c-\frac{b^2}{4a})\) Where ax^2+bx+c=0 is the standard form -4.9t^2 + 20t + 65=0 a=-4.9 b=20 c=65

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