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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?
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you just need the answers? If you need the work shown, are you doing differentiation and derivatives to do this or other methods?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i need the work shown and the ans please.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and i think were usin the quad formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0xcoordinate 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 ycoordinate

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or t=2 in this case XD

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0h=4.9(2)^2+19.6(2)+104 =19.6+39.2+104 =123.6 (2, 123.6)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0SO, the height is 123.6 and the time it took to get there is 2 seconds.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's the answer for the first two questions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks :)  from charmi too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0welcome, now, for the third question right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but like we learned in school , that the "b" value is the maximum height thats the part we dont get :S

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think that's right. The only way that would be true is if the coefficient of x^2 was 1/2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh, so like the question isnt right? for part (c) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0?? In question C, you just need to let h=0
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