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.(Quadratic Word Problems (Profit/Gravity)
y=-16x^2+121x+83

Answer 1

Answer:

The rocket will hit the ground after about 8.20 seconds.

Step-by-step explanation:

The height of the rocket y, in feet, x seconds after launch is modeled by the equation:

$y=-16x^2+121x+83$

We want to find the time at which the rocket will hit the ground.

If it hits the ground, the height of the rocket y will be 0. Thus:

$0=-16x^2+121x+83$

We can solve for x. Factoring (if possible at all) or completing the square can be tedious, so we can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = -16, b = 121, and c = 83. Substitute:

$\displaystyle x=\frac{-(121)\pm\sqrt{(121)^2-4(-16)(83)}}{2(-16)}$

Simplify:

$\displaystyle x=\frac{-121\pm\sqrt{19953}}{-32}$

Divide everything by -1 and simplify the square root. The plus/minus will remain unchanged:

$\displaystyle x=\frac{121\pm3\sqrt{2217}}{32}$

Therefore, our two solutions are:

$\displaystyle x=\frac{121+3\sqrt{2217}}{32}\approx 8.20\text{ or } x=\frac{121-3\sqrt{2217}}{32}\approx -0.63$

Since time cannot be negative, we can ignore the second solution.

Therefore, the rocket will hit the ground after about 8.20 seconds.