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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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?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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help please.
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
r yu sure? isnt the max height 104?

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Other answers:

no, the 104 is just the y-intercept, it says the object is thrown UP from that height |dw:1327885528660:dw|
ok hm how abt the rest of the ques
well the next one is 2 seconds because that was our x-coordinate of the vertex
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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