question\na rocket is launched from a tower. the height of the rocket, y in feet, is related to the time…

question\na 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.\n$y = -16x^2 + 89x + 50$\nanswer\nattempt 1 out of 5\nsubmit answer

question\na 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.\n$y = -16x^2 + 89x + 50$\nanswer\nattempt 1 out of 5\nsubmit answer

Answer

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 )