Question

A rocket is launched from a tower. The height of the rocket, y in feet, isrelated to the time after launch, x in seconds, by the given equation. Usingthis equation, find the time that the rocket will hit the ground, to the nearest100th of second.y = -16x^2+157x + 124

Answer 1

When x = 0, y = 124ft, which is the height of the tower.

They ground is y= 0. So, we have to solve the quadratic equation

-16x^2+157x + 124 = 0

Using Bhaskara:

Delta = 157² - 4(-16)(124)

Delta = 24649 + 7936

Delta = 32585

Solving the equation:

x1 = (-b +sqrt 32585)/2(-16)

x1 = (-157 + 180,513)/ -32

x1 = 23,513 / -32

x1 = -0,734 (there's no such thing as negative time)

x2= (-b -sqrt 32585)/2(-16)

x2 = (-157 - 180,513) -32

x2 = 10,547

Answer rounded to nearest 10th: x2 = 10,55s