72.0k views
1 vote
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= -16x2 + 1812 + 59

User MZB
by
7.9k points

1 Answer

1 vote

Answer:

The rocket will hit the ground after 113.28 seconds.

Explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:


ax^(2) + bx + c, a\\eq0.

This polynomial has roots
x_(1), x_(2) such that
ax^(2) + bx + c = a(x - x_(1))*(x - x_(2)), given by the following formulas:


x_(1) = (-b + √(\Delta))/(2*a)


x_(2) = (-b - √(\Delta))/(2*a)


\Delta = b^(2) - 4ac

The height of a rocket, after t seconds, is given by:


h(x) = -16x^2 + 1812x + 59

Using this equation, find the time that the rocket will hit the ground.

This is x for which
h(x) = 0. So


-16x^2 + 1812x + 59 = 0

Then
a = -16, b = 1812, c = 59


\Delta = (1812)^2 - 4(-16)(59) = 3287120


x_(1) = (-1812 + √(3287120))/(2*(-16)) = -0.03


x_(2) = (-1812 - √(3287120))/(2*(-16)) = 113.28

The rocket will hit the ground after 113.28 seconds.

User Ghostrifle
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories