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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the
time after launch, x in seconds, by the given equation. Using this equation, find the
time that the rocket will hit the ground, to the nearest 100th of second.
y= -16x2 + 1812 + 59

Sagot :

Answer:

The rocket will hit the ground after 113.28 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 a rocket, after t seconds, is given by:

[tex]h(x) = -16x^2 + 1812x + 59[/tex]

Using this equation, find the time that the rocket will hit the ground.

This is x for which [tex]h(x) = 0[/tex]. So

[tex]-16x^2 + 1812x + 59 = 0[/tex]

Then [tex]a = -16, b = 1812, c = 59[/tex]

[tex]\Delta = (1812)^2 - 4(-16)(59) = 3287120[/tex]

[tex]x_{1} = \frac{-1812 + \sqrt{3287120}}{2*(-16)} = -0.03[/tex]

[tex]x_{2} = \frac{-1812 - \sqrt{3287120}}{2*(-16)} = 113.28[/tex]

The rocket will hit the ground after 113.28 seconds.