anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
help please.
anonymous
  • anonymous
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
anonymous
  • anonymous
r yu sure? isnt the max height 104?

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