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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attempt 1 out of 5
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Explanation:

Step1: Set height to 0

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 get \( 16x^{2}-89x - 50 = 0 \).

Step2: Use quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \), the solutions are \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). 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 \( \sqrt{D}=\sqrt{11121}=105.456\) (approx). Now, \( x=\frac{89\pm105.456}{32} \). We have two solutions: \( x_1=\frac{89 + 105.456}{32}=\frac{194.456}{32}\approx6.07675 \) and \( x_2=\frac{89-105.456}{32}=\frac{-16.456}{32}\approx - 0.51425 \). Since time can't be negative, we take the positive solution.

Step3: Round to nearest 100th

Rounding \( 6.07675 \) to the nearest hundredth gives \( 6.08 \).

Answer:

\( 6.08 \)