Respuesta :
Answer:
The ball height is 11 meters at t = 0.59 seconds and t = 3.41 seconds.
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
The height of the ball after t seconds is given by:
[tex]h(t) = 1 + 20t - 5t^2[/tex]
Find all values of t for which the ball's height is 11 meters.
This is t for which [tex]h(t) = 11[/tex]. So
[tex]h(t) = 1 + 20t - 5t^2[/tex]
[tex]11 = 1 + 20t - 5t^2[/tex]
[tex]5t^2 - 20t + 10 = 0[/tex]
Simplifying by 5
[tex]t^2 - 4t + 2 = 0[/tex]
Which means that [tex]a = 1, b = -4, c = 2[/tex]
[tex]\Delta = (-4)^2 - 4(1)(2) = 8[/tex]
[tex]t_{1} = \frac{-(-4) + \sqrt{8}}{2} = 3.41[/tex]
[tex]t_{2} = \frac{-(-4) - \sqrt{8}}{2} = 0.59[/tex]
The ball height is 11 meters at t = 0.59 seconds and t = 3.41 seconds.
