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anonymous

  • 4 years ago

The top of a building in Ottawa is 104m above the ground. Suppose an object was thrown upward with an initial velocity of 19.6m/s from this height. The approximate height of the object above the ground, h metres, t seconds after being thrown, would be given by the function: h=-4.9t^2+19.6t+104 a) what was the maximum height of the object,above the ground? b) after how many seconds did the object reach its maximum height? c) From the time the object was initially thrown, how much did it takes to reach the ground, to the nearest second?

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  1. anonymous
    • 4 years ago
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    help please.

  2. anonymous
    • 4 years ago
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    we find the vertex for part a by first finding the axis of symmetry (-b/2a) which equals -19.6/-9.8 = 2. Now that is the x-value of the vertex so we plug it into x to get the y-value for that particular x-value. we get 123.6. The vertex is (2,123.6) Which means 123.6 is the maximum height

  3. anonymous
    • 4 years ago
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    r yu sure? isnt the max height 104?

  4. anonymous
    • 4 years ago
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    no, the 104 is just the y-intercept, it says the object is thrown UP from that height |dw:1327885528660:dw|

  5. anonymous
    • 4 years ago
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    ok hm how abt the rest of the ques

  6. anonymous
    • 4 years ago
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    well the next one is 2 seconds because that was our x-coordinate of the vertex

  7. anonymous
    • 4 years ago
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    for the last one we set y = 0 because at the ground there is no height so y = 0. Then we can use the quadratic formula to solve for x and it takes just over 7 seconds to reach the ground (7.02)

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