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babydoll332

  • 3 years ago

A projectile is thrown upward so that its distance above the ground after t seconds is h = - 16t2 + 440t. After how many seconds does it reach its maximum height?

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  1. Outkast3r09
    • 3 years ago
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    few ways to solve this

  2. completeidiot
    • 3 years ago
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    are you familiar with derivatives?

  3. babydoll332
    • 3 years ago
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    with a graph correct?

  4. Outkast3r09
    • 3 years ago
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    Without Calculus y|dw:1349657074734:dw|ou know that this graph will look like this

  5. babydoll332
    • 3 years ago
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    oh ok

  6. Outkast3r09
    • 3 years ago
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    so you know it hits its maximum height at its' vertex. you know how to find the x for the vertex by \[x=\frac{-b}{2a}\]

  7. Outkast3r09
    • 3 years ago
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    x so happens to be time in our case =P

  8. babydoll332
    • 3 years ago
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    ohhhh

  9. Outkast3r09
    • 3 years ago
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    |dw:1349657191197:dw|

  10. babydoll332
    • 3 years ago
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    21 seconds

  11. Outkast3r09
    • 3 years ago
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    with calculus \[y=-16t^2+440t\] \[y'=-32t+440=velocity\] we know that the velocity of y is 0 at the point of max height so \[0=-32t+440\] \[-440=-32t\] \[\frac{-440}{-32}=t\] 13.75 seconds

  12. Outkast3r09
    • 3 years ago
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    without calculus \[ax^2+bx+c\] in our case \[ax^2+bx\] \[x=\frac{-b}{2a}\] \[\frac{-440}{-32}\] 13.75 seconds

  13. babydoll332
    • 3 years ago
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    oh wow i was completly of thanks alot !!!

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