Question

the height of an arrow shot upward can be given by the formula ( s = v_0t - 16t^2 ), where ( v_0 ) is the initial velocity and ( t ) is time. how long does it take for the arrow to reach a height of 48 ft if it has an initial velocity of 96 ft/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
( (t - 6)^2 = 12 )
( (t - 3)^2 = 6 )
( (t + 9)^2 = -3 )

Explanation:

Step1: Start with the equation \(16t^2 - 96t + 48 = 0\)

Divide the entire equation by 16 to simplify: \(\frac{16t^2}{16}-\frac{96t}{16}+\frac{48}{16}=0\), which gives \(t^2 - 6t + 3 = 0\).

Step2: Move the constant term to the right side

Subtract 3 from both sides: \(t^2 - 6t = - 3\).

Step3: Complete the square

Take half of the coefficient of \(t\) (which is -6), so \(\frac{-6}{2}=-3\), square it: \((-3)^2 = 9\). Add 9 to both sides of the equation: \(t^2 - 6t + 9 = - 3 + 9\).

Step4: Rewrite the left side as a perfect square

The left side is \((t - 3)^2\), and the right side is 6. So \((t - 3)^2 = 6\).

Now, to solve for \(t\) from \((t - 3)^2 = 6\):

Step5: Take the square root of both sides

\(t - 3=\pm\sqrt{6}\)

Step6: Solve for \(t\)

\(t = 3\pm\sqrt{6}\)
Calculate the numerical values: \(\sqrt{6}\approx2.45\), so \(t_1 = 3 + 2.45 = 5.45\) and \(t_2 = 3 - 2.45 = 0.55\) (we consider the positive time values, and since the arrow is shot upward, both times are valid in context, but we are asked for when it reaches 48 ft, and the smaller time is on the way up, larger on the way down. If we just solve the quadratic for the given problem of reaching 48 ft with initial velocity 96 ft/s, we can also use the other equation \(48 = 96t - 16t^2\) which rearranges to \(16t^2 - 96t + 48 = 0\) as before. But from the completing the square we had \((t - 3)^2 = 6\), and if we were to match the given options, the completed square form is \((t - 3)^2 = 6\) (which is one of the options shown: \((t - 3)^2 = 6\)).

Answer:

The completed square form is \(\boldsymbol{(t - 3)^2 = 6}\), and the solutions for \(t\) (time to reach 48 ft) are approximately \(t\approx0.55\) seconds (on the way up) and \(t\approx5.45\) seconds (on the way down). If just the completed square form is needed, it's \((t - 3)^2 = 6\).