Respuesta :
Answer:
4 seconds, 256 ft
Step-by-step explanation:
First, we'll go through what needs to be done. Then we'll do it.
When h is 0, the fireworks is at a height of 0, or ground level. That happens before it is launched, at time 0 and when it falls back to the ground after going up and down. We let h = 0, and solve for t. We get t = 0, and another value for t which is when it hits the ground on the way down. The midpoint between the two times of height zero is the time at which maximum height is reached. Then we input the maximum height time into the equation and find h, the maximum height.
Now we'll solve it. First, we set the height equation equal to zero and solve for t.
-16t^2 + 128t = 0
-16t(t - 8) = 0
t = 0 or t - 8 = 0
t = 0 or t = 8
The height is zero at t = 0 seconds and at t = 8 seconds.
Maximum height is reached at the midpoint of the two times above.
(0 + 8)/2 = 8/2 = 4.
The firework explodes at 4 seconds.
Now we find the height at 4 seconds which is the maximum height.
h = -16t^2 + 128t
h = -16(4^2) + 128(4)
h = -256 + 512
h = 256
The firework explodes at a height of 256 ft.
This is about word problems of quadratic functions.
Height at which it explodes is 256 ft
- We are given the function;
h = -16t² + 128t
where;
h is height after time of t seconds
- At height of zero, the firework would either be about to launch or when it has come down.
Thus, let us set h = 0 to find the times.
0 = -16t² + 128t
Factorizing this gives us;
0 = -16t(t - 8)
Thus; -16t = 0 or t - 8 = 0
Thus, t = 0 secs or 8 secs
- Now the firework explodes at the highest point which will be the mid point of where the launching started and the endpoint where it stopped.
This means time at midpoint = (0 + 8)/2 = 4 secs
It explodes after 4 seconds.
Thus height at this time is;
h = -16(4)² + 128(4)
h = 256 ft
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