Question

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 a second.
f(x) = -16x^2+272x+110

Answer 1

Answer:

The time that the rocket will hit the ground will be:

• x = 17.40 seconds

Step-by-step explanation:

Given the expression

$f\left(x\right)\:=-16x^2+272x+110$

Here:

• x represents the time in seconds

• y represents the height of the rocket

We know that when the rocket will hit the ground, the height will get y = 0

i.e.

$0\:=-16x^2+272x+110$

solving for x to determine the time that the rocket takes when it will hit the ground.

so

$-16x^2+272x+110=0$

subtract 110 from both sides

$-16x^2+272x+110-110=0-110$

$-16x^2+272x=-110$

Divide both sides by -16

$\frac{-16x^2+272x}{-16}=\frac{-110}{-16}$

$x^2-17x=\frac{55}{8}$

$\mathrm{Add\:}\left(-\frac{17}{2}\right)^2\mathrm{\:to\:both\:sides}$

$x^2-17x+\left(-\frac{17}{2}\right)^2=\frac{55}{8}+\left(-\frac{17}{2}\right)^2$

$x^2-17x+\left(-\frac{17}{2}\right)^2=\frac{633}{8}$

$\left(x-\frac{17}{2}\right)^2=\frac{633}{8}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{17}{2}=\sqrt{\frac{633}{8}}$

$x=\frac{\sqrt{1266}}{4}+\frac{17}{2}$

x = 17.40 seconds

also solving

$x-\frac{17}{2}=-\sqrt{\frac{633}{8}}$

$x=-\frac{\sqrt{1266}}{4}+\frac{17}{2}$

x = -0.40 seconds

As time can not be negative.

Thus, the value of x = 17.40 seconds

Therefore, the time that the rocket will hit the ground will be:

• x = 17.40 seconds