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+232x+134

Answer 1

Explanation:

when it touch the ground,y=0

-16x²+232x+134=0

-16(x²-29/2 x+(-29/4)²-(-29/4)²)=-134

-16(x-29/4)²-16(-861/16)=-134

-16(x-29/4)²+861=-134

-16(x-29/4)²=-134-861

(x-29/4)²=-995/-16=995/16

x-29/4=±√995/4

rejecting negative sign

x=29/4+√995/4=(29+√995)/4≈15.14 second