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.

$y = -16x^{2}+141x + 67$

Explanation:

Step1: Set height to 0

When the rocket hits the ground, $y = 0$. So we have the quadratic equation $-16x^{2}+141x + 67=0$.

Step2: Identify coefficients

For the quadratic equation $ax^{2}+bx + c = 0$, here $a=-16$, $b = 141$, $c = 67$.

Step3: Use quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute the values: $x=\frac{-141\pm\sqrt{141^{2}-4\times(-16)\times67}}{2\times(-16)}$.

Step4: Calculate discriminant

First, calculate the discriminant $\Delta=b^{2}-4ac=141^{2}-4\times(-16)\times67=19881 + 4288=24169$.

Step5: Find two solutions

$x=\frac{-141\pm\sqrt{24169}}{-32}=\frac{-141\pm155.4766}{-32}$.
We get two solutions:
$x_1=\frac{-141 + 155.4766}{-32}=\frac{14.4766}{-32}\approx - 0.45$ (rejected since time can't be negative).
$x_2=\frac{-141-155.4766}{-32}=\frac{-296.4766}{-32}\approx9.20$.

Answer:

$9.20$ seconds