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+110x+77

Answer 1

-16x² + 110x + 77 = 0

16x² - 110x - 77 = 0

x = (110 ± √((-110)² - 4(16)(-77)))/32

= (110 + √17,028)/32

= (110 + 6√473)/32

= (55 + 3√473)/16

= about 7.52 seconds