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+ 181x + 59

Answer 1

The rocket will hit the ground after about 11.63 seconds.

Explanation:

A rocket is launched from a tower. Its height y in feet after x seconds is modeled by the equation:

`y=-16x^2+181x+59`

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

If it hits the ground, its height y above the ground will be 0. So, we can set y = 0 and solve for x:

`0=-16x^2+181x+59`

Solve the quadratic. Factoring is too tedious or impossible, and completing the square is also very tedious. So, we can use the quadratic formula:

`\displaystyle x=(-b\pm√(b^2-4ac))/(2a)`

In this case, a = -16, b = 181, and c = 59. Substitute:

`\displaystyle x=(-(181)\pm√((181)^2-4(-16)(59)))/(2(-16))`

Evaluate:

`\displaystyle x=(-181\pm√(36537))/(-32)`

Simplify:

`\displaystyle x=(181\pm√(36537))/(32)`

Hence, our solutions are:

`\displaystyle x=(181+√(36537))/(32)\approx 11.63\text{ or } x=(181-√(36537))/(32)\approx-0.32`

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

So, the rocket will hit the ground after about 11.63 seconds.