A ball is dropped from a diff that is 320 feet high. The distance \( S \) (infeet) that it falls in \( t \) seconds is given by the formula \( s=16 t^{2} \). How many seconds (to tenths) will it take for the ball to hit the graind?

Given: - The height from which the ball is dropped is 320 feet. - The distance \( S \) (in feet) that the ball falls in \( t \) seconds is given by the formula \( S = 16t^2 \). To find: - The time it takes for the ball to hit the ground. We know that when the ball hits the ground, the distance it has fallen is equal to the height from which it was dropped. So, we can set up the equation: \[ 320 = 16t^2 \] Now, we can solve this equation to find the value of \( t \), which represents the time it takes for the ball to hit the ground. Solve the quadratic equation by following steps: - step0: Solve using square roots: \(320=16t^{2}\) - step1: Swap the sides: \(16t^{2}=320\) - step2: Divide both sides: \(\frac{16t^{2}}{16}=\frac{320}{16}\) - step3: Divide the numbers: \(t^{2}=20\) - step4: Simplify the expression: \(t=\pm \sqrt{20}\) - step5: Simplify the expression: \(t=\pm 2\sqrt{5}\) - step6: Separate into possible cases: \(\begin{align}&t=2\sqrt{5}\\&t=-2\sqrt{5}\end{align}\) - step7: Rewrite: \(t_{1}=-2\sqrt{5},t_{2}=2\sqrt{5}\) The solutions to the equation \(320 = 16t^2\) are \(t = -2\sqrt{5}\) and \(t = 2\sqrt{5}\). Since time cannot be negative, the time it takes for the ball to hit the ground is \(t = 2\sqrt{5}\) seconds. To find the time in seconds to the nearest tenth, we can calculate \(2\sqrt{5}\) and round it to the nearest tenth. Calculate the value by following steps: - step0: Calculate: \(2\sqrt{5}\) It will take approximately 4.5 seconds for the ball to hit the ground.