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+157x+124 y=−16x 2 +157x+124

Answer 1

Rocket will hit the ground in 10.55 seconds.

Explanation:

Let be
`y = -16\cdot x^(2)+157\cdot x +124`, where
`x` and
`y` are time after launch and the height of the rocket, measured in seconds and meters, respectively. To find the time that the rocket will hit the ground, we need to equalize height to zero and solve the resulting expression:

`-16\cdot x^(2)+157\cdot x + 124 = 0`

The quadratic function can be solved analitically by Quadratic Formula, that is to say:

`x = \frac{-157\pm \sqrt{157^(2)-4\cdot (-16)\cdot (124)}}{2\cdot (-16)}`

Roots of the polynomial are, respectively:

`x_(1) \approx 10.55\,s`,
`x_(2)\approx -0.74\,s`

Only first root offers to us a reasonable solution. Hence, we conclude that rocket will hit the ground in 10.55 seconds.