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.

$y = -16x^{2}+172x + 146$

Explanation:

Step1: Set height to 0

When the rocket hits the ground, $y = 0$. So we have the quadratic equation $-16x^{2}+172x + 146=0$. Divide through by -2 to simplify: $8x^{2}-86x - 73 = 0$.

Step2: Use quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 8$, $b=-86$, and $c=-73$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-86)^{2}-4\times8\times(-73)=7396 + 2336=9732$.

Step3: Find the values of x

$x=\frac{86\pm\sqrt{9732}}{16}=\frac{86\pm98.65}{16}$. We have two solutions: $x_1=\frac{86 + 98.65}{16}=\frac{184.65}{16}\approx11.54$ and $x_2=\frac{86-98.65}{16}=\frac{-12.65}{16}\approx - 0.79$. Since time cannot be negative, we discard the negative - valued solution.

Answer:

$11.54$