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 second.
y=-16x^2+165x+69

Answer 1

Answer:

The rocket hits the gorund after approximately 10.71 seconds.

Step-by-step explanation:

The height of the rocket y in feet x seconds after launch is given by the equation:

$y=-16x^2+165x+69$

And we want to find the time in which the rocket will hit the ground.

When it hits the ground, its height above ground will be 0. Hence, we can let y = 0 and solve for x:

$0=-16x^2+165x+69$

We can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = -16, b = 165, and c = 69.

Substitute:

$\displaystyle x=\frac{-165\pm\sqrt{(165)^2-4(-16)(69)}}{2(-16)}$

Evaluate:

$\displaystyle x=\frac{-165\pm\sqrt{31641}}{-32}=\frac{165\pm\sqrt{31641}}{32}$

Hence, our solutions are:

$\displaystyle x_1=\frac{165+\sqrt{31641}}{32}\approx 10.71\text{ or } x_2=\frac{165-\sqrt{31641}}{32}\approx-0.40$

Since time cannot be negative, we can ignore the first answer.

So, the rocket hits the gorund after approximately 10.71 seconds.