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

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.\n\n$y = -16x^{2}+273x + 150$

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.\n\n$y = -16x^{2}+273x + 150$

Answer

Explanation:

Step1: Set height y to 0

When the rocket hits the ground, $y = 0$. So we have the quadratic equation $-16x^{2}+273x + 150=0$. For a quadratic equation $ax^{2}+bx + c = 0$ (here $a=-16$, $b = 273$, $c = 150$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Step2: Calculate the discriminant

First, calculate the discriminant $\Delta=b^{2}-4ac=(273)^{2}-4\times(-16)\times150=74529 + 9600=84129$.

Step3: Find the values of x

$x=\frac{-273\pm\sqrt{84129}}{2\times(-16)}=\frac{-273\pm290.05}{-32}$. We get two solutions: $x_1=\frac{-273 + 290.05}{-32}=\frac{17.05}{-32}\approx - 0.53$ (rejected since time cannot be negative). $x_2=\frac{-273-290.05}{-32}=\frac{-563.05}{-32}\approx17.59$.

Answer:

$17.59$