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

Explanation:

Step1: Divide equation by 16

Divide $16t^{2}-96t + 48=0$ by 16 to simplify.
$t^{2}-6t + 3=0$

Step2: Rearrange terms

Move the constant term to the right - hand side.
$t^{2}-6t=-3$

Step3: Complete the square

Take half of the coefficient of $t$, square it and add to both sides. The coefficient of $t$ is - 6. Half of it is - 3, and its square is 9.
$t^{2}-6t + 9=-3 + 9$
$(t - 3)^{2}=6$

Step4: Solve for t

Take the square - root of both sides.
$t-3=\pm\sqrt{6}$
$t = 3\pm\sqrt{6}$
$t_1=3+\sqrt{6}\approx3 + 2.45=5.45$
$t_2=3-\sqrt{6}\approx3 - 2.45 = 0.55$

Answer:

$(t - 3)^{2}=6$