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 (\boxed{}) (options: ( (t + 9)^2 = -3 ), ( (t - 3)^2 = 6 ), ( (t - 6)^2 = 12 ))

Explanation:

Step1: Simplify the quadratic equation

Divide all terms by 16:
$$\frac{16t^2}{16} - \frac{96t}{16} + \frac{48}{16} = 0$$
$$t^2 - 6t + 3 = 0$$

Step2: Rearrange the equation

Isolate the constant term:
$$t^2 - 6t = -3$$

Step3: Complete the square

Add $(\frac{-6}{2})^2=9$ to both sides:
$$t^2 - 6t + 9 = -3 + 9$$
$$(t - 3)^2 = 6$$

Answer:

$(t - 3)^2 = 6$