Question

a rock is thrown from the top of a tall building. the distance, in feet, between the rock and the ground x seconds after it is thrown is given by $f(x) = -16x^2 - 4x + 382$ how long after the rock is thrown does it hit the ground?

Explanation:

Step1: Set distance to 0 (ground hit)

$f(x) = 0 \implies -16x^2 -4x + 382 = 0$

Step2: Simplify the quadratic equation

Multiply by -1: $16x^2 +4x - 382 = 0$
Divide by 2: $8x^2 +2x - 191 = 0$

Step3: Apply quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here $a=8$, $b=2$, $c=-191$

$$ x=\frac{-2\pm\sqrt{2^2-4(8)(-191)}}{2(8)} $$

Step4: Calculate discriminant

$\sqrt{4 + 6112} = \sqrt{6116} \approx 78.20$

Step5: Solve for positive x

$x=\frac{-2 + 78.20}{16} \approx \frac{76.20}{16} \approx 4.76$
(Disregard negative root as time can't be negative)

Answer:

Approximately 4.76 seconds