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

Answer:

Time required to hit the ground is 7.9 s.

Step-by-step 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

$-16x^2+119x+57=0\\\\16 x^2 - 119 x - 57 = 0 \\\\x = \frac{119\pm\sqrt{14161+3648}}{32}\\\\x = \frac{119\pm133.45}{32}\\\\t = - 0.45 s, 7.9 s$

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