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. Find the time that the rocket will hit the ground, to the nearest 100th of second. Y=-16x^2+189x+79

Answer 1

Step-by-step explanation:

Given the height of the rocket y in feet related to time after launch x related by the equation:

Y=-16x^2+189x+79

The rocket will hit the ground when y = 0

The equation will become:

0=-16x^2+189x+79

Rearrange

-16x^2+189x+79 = 0

multiply though by minus

16x^2-189x-79 = 0

Factorize and solve for x:

x = 189±√189²-4(16)(-79)/2(16)

x = 189±√35721+5056/32

x = 189±√40777/32

x = 189±201.93/32

x = 189+201.93/32

x = 390.93/32

x = 12.22secs

hence the rocket will hit the ground after 12.22seconds (to the nearest hundredth of a sec)