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.

Answer 1

Here is the full 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.

Equation:

y=-16x^2+153x+98

Answer:

10.2 seconds

Explanation:

Given equation :

`y=-16x^2+153x+98` shown the expression of a quadratic equation

Let y =0

∴

0 =
`-16x^2+153x+98`

where;

`a=-16\\b=153\\c=98`

Using the quadratic formula:

`x=(-b\±√(b^2-4ac))/(2a)`

replacing it with our values; we have:

`x=(-153\±√(153^2-4(-16)(98)))/(2(-16))`

`x=(-153\ + √(153^2-4(-16)(98)))/(2(-16))` OR
`x=(-153\ - √(153^2-4(-16)(98)))/(2(-16))`

`x_1=10.17` OR
`x_2=-0.60`

Hence, we go by the positive value since, is the time that the rocket will hit the ground.

x= 10.17

x = ≅ 10.2 seconds

Therefore, the rocket will hit the ground, to the nearest 100th of second. = 10.2 seconds