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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.

equation: y=−16x^2+177x+98

User Roei
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1 Answer

11 votes

Answer:

The rocket hits the ground at a time of 11.59 seconds.

Explanation:

The height of the rocket, after x seconds, is given by the following equation:


y = -16x^2 + 177x + 98

It hits the ground when
y = 0, so we have to find x for which
y = 0, which is a quadratic equation.

Finding the roots of 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 + √(\bigtriangleup))/(2*a)


x_(2) = (-b - √(\bigtriangleup))/(2*a)


\bigtriangleup = b^(2) - 4ac

In this question:


y = -16x^2 + 177x + 98


-16x^2 + 177x + 98 = 0

So


a = -16, b = 177, c = 98


\bigtriangleup = 177^(2) - 4(-16)(98) = 37601


x_(1) = (-177 + √(37601))/(2*(-16)) = -0.53


x_(2) = (-177 - √(37601))/(2*(-16)) = 11.59

Since time is a positive measure, the rocket hits the ground at a time of 11.59 seconds.

User Lipsumar
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