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.

$y = - 16x^{2}+258x + 97$

Explanation:

Step1: Set height to 0

When the rocket hits the ground, $y = 0$. So we have the quadratic equation $-16x^{2}+258x + 97=0$.

Step2: Identify coefficients

For a quadratic equation $ax^{2}+bx + c = 0$, here $a=-16$, $b = 258$, $c = 97$.

Step3: Use quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute the values: $x=\frac{-258\pm\sqrt{258^{2}-4\times(-16)\times97}}{2\times(-16)}$.

Step4: Calculate discriminant

First, calculate the discriminant $\Delta=b^{2}-4ac=258^{2}-4\times(-16)\times97=66564 + 6208=72772$.

Step5: Find x - values

$x=\frac{-258\pm\sqrt{72772}}{-32}$. $\sqrt{72772}\approx269.763$. So $x=\frac{-258\pm269.763}{-32}$.
We get two solutions: $x_1=\frac{-258 + 269.763}{-32}=\frac{11.763}{-32}\approx - 0.37$ and $x_2=\frac{-258-269.763}{-32}=\frac{-527.763}{-32}\approx16.49$.
Since time cannot be negative, we discard the negative solution.

Answer:

$16.49$