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

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 = -16x² + 273x + 111

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 = -16x² + 273x + 111

Answer

Explanation:

Step1: Set height y to 0

When the rocket hits the ground, $y = 0$. So we have the quadratic - equation $-16x^{2}+273x + 111=0$. The general form of a quadratic equation is $ax^{2}+bx + c = 0$, where $a=-16$, $b = 273$, and $c = 111$.

Step2: Use the quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(273)^{2}-4\times(-16)\times111$. $=74529+7104=81633$. Then, $x=\frac{-273\pm\sqrt{81633}}{2\times(-16)}=\frac{-273\pm285.715}{-32}$.

Step3: Find the two solutions for x

We have two solutions: $x_1=\frac{-273 + 285.715}{-32}=\frac{12.715}{-32}\approx - 0.397$ (rejected since time cannot be negative). $x_2=\frac{-273-285.715}{-32}=\frac{-558.715}{-32}\approx17.46$.

Answer:

$17.46$