What building?
And where does the balloon come in ?
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
3 answers
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)
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)