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