Question

1. an object is propelled vertically upward from the top of a 16 - foot building. the quadratic function ( s(t)=-16t^{2}+128t + 16 ) models the balls height above the ground, ( s(t) ), in feet, ( t ) seconds after it was thrown. how many seconds does it take until the object finally hits the ground? round to the nearest tenth of a second if necessary.

Explanation:

Step1: Set up the equation

When the object hits the ground, \(s(t) = 0\). So we have the equation \(-16t^{2}+128t + 16=0\). Divide through by \(-16\) to simplify: \(t^{2}-8t - 1=0\).

Step2: Apply the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-8\), \(c=-1\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute the values: \(t=\frac{8\pm\sqrt{(-8)^{2}-4\times1\times(-1)}}{2\times1}=\frac{8\pm\sqrt{64 + 4}}{2}=\frac{8\pm\sqrt{68}}{2}=\frac{8\pm2\sqrt{17}}{2}=4\pm\sqrt{17}\).

Step3: Determine the valid solution

We have two solutions for \(t\): \(t = 4+\sqrt{17}\approx4 + 4.123=8.1\) and \(t=4-\sqrt{17}\approx4-4.123=- 0.1\). Since time \(t\geq0\), we discard the negative solution.

Answer:

\(t\approx8.1\) seconds