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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+281x+137

1 Answer

1 vote

Answer:

It will take 18.04s for the rocket to hit the ground.

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 = (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 problem:

Height given by the following equation:


y = -16x^(2) + 281x + 137

It hits the ground when y = 0. So


-16x^(2) + 281x + 137 = 0

Multiplying by -1


16x^(2) - 281x - 137 = 0

So
a = 16, b = -281, c = -137

Then


\bigtriangleup = (-281)^(2) - 4*16*(-137) = 87729


t_(1) = (-(-281) + √(87729))/(2*16) = 18.04


t_(2) = (-(-281) - √(87729))/(2*16) = -0.4747

It cannot take negative time, so we discard
t_(2)

It will take 18.04s for the rocket to hit the ground.

User Nullglob
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