Respuesta :
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]
- The time at which the firework reaches the maximum height is t = 1 seconds
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
- The maximum height of the firework, f(1) = 25 yards
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;
- The time after which the firework will fall to the ground, t = 1 + √5 seconds
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