The height of an arrow shot upward can be given by the formula s=v_(0)t-16t^2 where v_(0) is the initial velocity and tis time How long does it take for the arrow to reach a height of 48 ft if it has an initial velocity of 96ft/s ? Round to the nearest hundredth. The equation that represents the problem is 48=96t-16t^2 Solve 16t^2-96t+48=0 Complete the square to write 16t^2-96t+48=0 as DONE V (t+9)^2=-3 (t-3)^2=6 (t-6)^2=12

## Step 1<br />The problem involves the physics concept of motion under constant acceleration, specifically the motion of an object thrown upwards. The formula given is \(s = v_{0}t - 16t^{2}\), where \(s\) is the height, \(v_{0}\) is the initial velocity, and \(t\) is the time. The problem asks us to find the time it takes for the arrow to reach a height of 48ft, given an initial velocity of 96ft/s.<br /><br />## Step 2<br />We start by setting up the equation that represents the problem, which is \(48 = 96t - 16t^{2}\). This equation is derived from the given formula by substituting the given values for \(s\) and \(v_{0}\).<br /><br />## Step 3<br />Next, we rearrange the equation to the standard quadratic form \(at^{2} + bt + c = 0\), which gives us \(16t^{2} - 96t + 48 = 0\).<br /><br />## Step 4<br />We then solve this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^{2} - 2a}\). Substituting the values of \(a\), \(b\), and \(c\) from our equation, we get \(t = \frac{96 \pm \sqrt{(-96)^{2} - 4*16*48}}{2*16}\).<br /><br />## Step 5<br />Solving this gives us two solutions, \(t = 3\) and \(t = 6\). However, since time cannot be negative, we only consider the positive root, which is \(t = 3\).