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+147x+80

Answer 1

Answer:

t = 9.7 seconds

Step-by-step explanation:

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 as :

$y=-16x^2+147x+80$

We need to find the time that the rocket will hit the ground, to the nearest 100th of second.

When the rocket hit the ground, So,

$-16x^2+147x+80=0\\\\x=-0.515\ s\ and\ 9.70\ s$

Neglecting negative tim,

t = 9.7 s

Hence, it will hit the ground at 9.7 seconds.