a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time after launch…

Answer:

$7.54$ seconds

Explanation:

Step1: Set height to 0

When the rocket hits the ground, $y = 0$. So we have the quadratic equation $-16x^{2}+110x + 77=0$.

Step2: Identify coefficients

For the quadratic equation $ax^{2}+bx + c = 0$, here $a=-16$, $b = 110$, $c = 77$.

Step3: Apply quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute the values: $x=\frac{-110\pm\sqrt{110^{2}-4\times(-16)\times77}}{2\times(-16)}$.

Step4: Calculate discriminant

First, calculate the discriminant $\Delta=b^{2}-4ac=110^{2}-4\times(-16)\times77=12100 + 4928=17028$.

Step5: Find x - values

$x=\frac{-110\pm\sqrt{17028}}{-32}=\frac{-110\pm130.49}{-32}$. We get two solutions: $x_1=\frac{-110 + 130.49}{-32}\approx - 0.64$ and $x_2=\frac{-110-130.49}{-32}\approx7.54$.

Step6: Select valid solution

Since time cannot be negative, we discard $x_1$. So the time is approximately $7.54$ seconds.