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
schrodinger
  • schrodinger
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anonymous
  • anonymous
a) 104 m
anonymous
  • anonymous
whats b
anonymous
  • anonymous
D: y is there 2 heights in the equation, theres like on (h) and then the max height (104)

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anonymous
  • anonymous
idk thts wht the ques is i found it frm our test paper
anonymous
  • anonymous
u have my test paper right -_-
anonymous
  • anonymous
yup
anonymous
  • anonymous
buh yu dint ans it
anonymous
  • anonymous
ohhh
anonymous
  • anonymous
u left it blank LOL
anonymous
  • anonymous
lmfaoo yee i dont know, ima try to do it
anonymous
  • anonymous
i posted it again scroll up
anonymous
  • anonymous
im gunna post it

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