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+143x+121
Answers
Answer:
9.72 seconds ( approx )
Step-by-step explanation:
Since, the equation that shows the height of the rocket from the ground,
Where,
x = time after launch, in seconds,
When the rocket hits the ground,
y = 0,
i.e.
By the quadratic formula,
∵ Time can not be negative,
So, the time taken to hit the ground = 9.716 seconds ≈ 9.72 seconds.
Answer:
9.72 seconds (approx)
Step-by-step explanation:
Since, the equation that shows the height of the rocket from the ground,
y=-16x^2+143x+121y=−16x
2
+143x+121
Where,
x = time after launch, in seconds,
When the rocket hits the ground,
y = 0,
i.e.
-16x^2 + 143x+12=0−16x
2
+143x+12=0
By the quadratic formula,
x = \frac{-143\pm \sqrt{(143)^2 - 4\times -16\times 12}}{-32}x=
−32
−143±
(143)
2
−4×−16×12
x = \frac{-143\pm \sqrt{20449 + 7744}}{-32}x=
−32
−143±
20449+7744
\implies x = -0.778\text{ or }x=9.716⟹x=−0.778 or x=9.716
∵ Time can not be negative,
So, the time taken to hit the ground = 9.716 seconds ≈ 9.72 seconds.