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katrin [286]
3 years ago
5

A firework is fired into the air from the top of a barn. Tis hight (h) above the ground in yards after (t) seconds is given by t

he function h(t)= -5t^2 + 10t + 20
1. At what time does the firework reach its maximum hight?
2. What is the maximum hight of the firework?
3. At what time will the firework fall to the ground?
Mathematics
1 answer:
Grace [21]3 years ago
3 0

The given function for the height of the firework is a quadratic function

1. Time at which the firework reaches the maximum height is <u>1 seconds</u>

2. The maximum height of the firework, is <u>25 yards</u>

3. Time after which the firework will fall to the ground is<u> (1 + √5) seconds</u>

<u />

Reason:

The given function that represents the height of the fireworks with time is

presented as follows;

h(t) = -5·t² + 10·t + 20

1. The time at which the firework reaches its maximum height is given by

the maximum point of the given function as follows;

The x-value of the maximum point of a quadratic function is x = \dfrac{b}{2 \cdot a}

Where;

<em>a</em>, and <em>b</em>, are the coefficient of x² and <em>x</em>, in the general form of a quadratic function f(x) = a·x² + b·x + c

By comparison, we have;

t = -\dfrac{10}{2 \times (-5)}  = 1

  • The time at which the firework reaches the maximum height is t = <u>1 seconds</u>

2. The maximum height is given by plugging in the value of <em>t</em>, at the maximum point into the given function as follows;

h(1) = -5×1² + 10×1 + 20 = 25

  • The maximum height of the firework, f(1) = <u>25 yards</u>

3 The time at which the firework will fall to the ground, is given by the zero of the function as follows;

When the firework falls to the ground, h(t) = 0 =  -5·t² + 10·t + 20

Dividing both sides by (-5) gives;

\dfrac{0}{-5} =  \dfrac{  -5 \cdot t^2 + 10 \cdot t + 20}{-5} = t^2 - 2 \cdot t - 4

t² - 2·t - 4 = 0

By the quadratic formula x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}, we get;

t = \dfrac{2\pm \sqrt{(-2)^{2}-4\times  1\times (-4)}}{2\times 1} = \dfrac{2\pm \sqrt{20}}{2\times 1} = \dfrac{2\pm 2 \times \sqrt{5}}{2\times 1}  = 1 \pm \sqrt{5}

Therefore;

  • The time after which the firework will fall to the ground, t = <u>1 + √5 seconds</u>

Learn more here:

brainly.com/question/20628403

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