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+119x+57

Answer 1

Time required to hit the ground is 7.9 s.

Explanation:

The height of the rocket is given by

`y =- 16 x^2 + 119 x + 57`

For the time to hit the ground, put y = 0

`-16x^2+119x+57=0\\\\16 x^2 - 119 x - 57 = 0 \\\\x = (119\pm√(14161+3648))/(32)\\\\x = (119\pm133.45)/(32)\\\\t = - 0.45 s, 7.9 s`

Time cannot be negative, so time to hit the ground is 7.9 s.