30) an object is propelled vertically upward from the top of a 16 - foot building. the quadratic function (…

30) 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.

30) 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.

Answer

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