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 + 1812 + 59

Answer 1

The rocket will hit the ground after 113.28 seconds.

Explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\Delta))/(2*a)`

`x_(2) = (-b - √(\Delta))/(2*a)`

`\Delta = b^(2) - 4ac`

The height of a rocket, after t seconds, is given by:

`h(x) = -16x^2 + 1812x + 59`

Using this equation, find the time that the rocket will hit the ground.

This is x for which
`h(x) = 0`. So

`-16x^2 + 1812x + 59 = 0`

Then
`a = -16, b = 1812, c = 59`

`\Delta = (1812)^2 - 4(-16)(59) = 3287120`

`x_(1) = (-1812 + √(3287120))/(2*(-16)) = -0.03`

`x_(2) = (-1812 - √(3287120))/(2*(-16)) = 113.28`

The rocket will hit the ground after 113.28 seconds.