Question

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.
$y = -16x^2 + 89x + 50$
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Explanation:

Step1: Set \( y = 0 \) (rocket hits ground)

We have the equation \( y=-16x^{2}+89x + 50 \). When the rocket hits the ground, \( y = 0 \), so we solve the quadratic equation \( -16x^{2}+89x + 50=0 \). Multiply both sides by -1 to make it easier: \( 16x^{2}-89x - 50 = 0 \).

Step2: Use quadratic formula

The quadratic formula for \( ax^{2}+bx + c = 0 \) is \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Here, \( a = 16 \), \( b=-89 \), \( c=-50 \). First, calculate the discriminant \( D=b^{2}-4ac=(-89)^{2}-4\times16\times(-50)=7921 + 3200 = 11121 \). Then, \( x=\frac{89\pm\sqrt{11121}}{32} \). Calculate \( \sqrt{11121}\approx105.456 \). So we have two solutions: \( x=\frac{89 + 105.456}{32}\approx\frac{194.456}{32}\approx6.07675 \) and \( x=\frac{89 - 105.456}{32}\approx\frac{-16.456}{32}\approx - 0.51425 \). Since time can't be negative, we take the positive solution.

Step3: Round to nearest hundredth

\( 6.07675\approx6.08 \) (to the nearest hundredth).

Answer:

\( 6.08 \)