Posted by **Vieanna** on Tuesday, July 2, 2013 at 8:37pm.

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

- Math -
**Reiny**, Tuesday, July 2, 2013 at 9:05pm
What building?

And where does the balloon come in ?

- Math -
**Vieanna**, Tuesday, July 2, 2013 at 9:08pm
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

- Math -
**Reiny**, Tuesday, July 2, 2013 at 9:56pm
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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