Math
posted by Vieanna .
the function is h(t)= 4.9t^2+2t+45 (t represents time in seconds and h represents height in metres.)
a) how tall is the building?
b) how long does it take the balloon to fall on the sidewalk?
c)determine the max height of balloon

What building?
And where does the balloon come in ? 
The question states that the water balloon is thrown from the roof of a building. the path that the balloon follows the following function, which was listed above

Ok,
a) so when the balloon was thrown from the building
t = 0
so h = 0 + 0 + 45
The building is 45 metres high
b) when it hits the ground, h = 0
0 = 4.9t^2 + 2t + 45
4.9t^2  2t  45 = 0
by the formula:
t = (2 ± √(4  4(4.9)(45))/9.8
= 3.24 or a negative, which we would reject
it took 3.24 seconds to hit the ground
c) If you know Calculus ....
dh/dt = 9.8t + 2 = 0 for a max height
9.8t = 2
t = .2041 seconds
at that time, h = 4.9(.2041^2) + 2(.2041) + 45
= 45.2041m
If you don't know Calculus, then you probably learned how to complete the square
h = 4.9(t^2 (2/4.9)t +.04165  .04165) + 45
= 4.9(t  .2041)^2 + 45.2041
So the vertex is (.2041 , 45.2041) > max is 45.2041
(same as above)
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