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 the 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?

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

1. Time at which the firework reaches the maximum height is 1 seconds

2. The maximum height of the firework, is 25 yards

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

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 [tex]x = \dfrac{b}{2 \cdot a}[/tex]

Where;

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

By comparison, we have;

[tex]t = -\dfrac{10}{2 \times (-5)} = 1[/tex]

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

h(1) = -5×1² + 10×1 + 20 = 25

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;

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

t² - 2·t - 4 = 0

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

[tex]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}[/tex]

Therefore;

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