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anonymous

  • one year ago

A rock is launched from a cannon. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. After 1 second, the rock is 129 feet in the air; after 2 seconds, it is 236 feet in the air. Find the height, in feet, of the rock after 5 seconds in the air.

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  1. welshfella
    • one year ago
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    we can use the established equation of motion under acceleration due to gravity( = 32 ft s-2 ) and the equation is s = ut + 0.5at^2 where u = initial velocity , a = acceleration = -32 , s = distance and t = time s = ut - 16t^2 so plugging in given values 129 = u(1) - 16(1)^2 129 = u - 16 u = 145 so our equation is s = 145t - 16t^2

  2. anonymous
    • one year ago
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    so whats the answer? haha sorry im really bad at thinking right now

  3. welshfella
    • one year ago
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    when t = 5 height s = 145*5 - 16*5^2 = 325 feet

  4. anonymous
    • one year ago
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    it was incorect

  5. welshfella
    • one year ago
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    h(x) after 5 seconds is 325 feet

  6. welshfella
    • one year ago
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    well thats the correct equation

  7. welshfella
    • one year ago
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    another value of u is 150 which will give a result of 350

  8. welshfella
    • one year ago
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    if i use the 236 ft after 2 seconds it gives u = 150

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