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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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Do you just need the answers? If you need the work shown, are you doing differentiation and derivatives to do this or other methods?
anonymous
  • anonymous
i need the work shown and the ans please.
anonymous
  • anonymous
and i think were usin the quad formula

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anonymous
  • anonymous
ohh, ok XD
anonymous
  • anonymous
x-coordinate of the Vertex or the maximum height is found using this equation: -b/2a a=-4.9 b=19.6 -19.6/(2(-4.9)) =-19.6/-9.8 =2 x=2 for vertex Plug it back in to figure out the y-coordinate
anonymous
  • anonymous
or t=2 in this case XD
anonymous
  • anonymous
h=-4.9(2)^2+19.6(2)+104 =-19.6+39.2+104 =123.6 (2, 123.6)
anonymous
  • anonymous
SO, the height is 123.6 and the time it took to get there is 2 seconds.
anonymous
  • anonymous
that's the answer for the first two questions
anonymous
  • anonymous
thanks :) - from charmi too
anonymous
  • anonymous
welcome, now, for the third question right?
anonymous
  • anonymous
but like we learned in school , that the "b" value is the maximum height thats the part we dont get :S
anonymous
  • anonymous
just a sec
anonymous
  • anonymous
I don't think that's right. The only way that would be true is if the coefficient of x^2 was 1/2.
anonymous
  • anonymous
ohh, so like the question isnt right? for part (c) ?
anonymous
  • anonymous
?? In question C, you just need to let h=0
anonymous
  • anonymous
ohh ok thanks :)

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