Question

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+199x+90

Answer 1

Answer:

The time that the rocket will hit the ground is 12.87 seconds.

Step-by-step explanation:

Given : The rocket is launched from a tower  $y=-16x^2+199x+90$. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation.

To find : The time that the rocket will hit the ground ?

Solution :

When the rocket hit the ground i.e. height became zero y=0,

Equation is $y=-16x^2+199x+90$

Substitute y=0,

$-16x^2+199x+90=0$

Using quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(199)\pm\sqrt{(199)^2-4(-16)(90)}}{2(-16)}$

$x=\frac{-199\pm\sqrt{45361}}{-32}$

$x=\frac{-199+\sqrt{45361}}{-32},\frac{-199-\sqrt{45361}}{-32}$

$x=-0.43,12.87$

Reject negative value.

The time taken is 12.87 seconds.