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: Rearrange the equation

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

Step2: Isolate constant term

Move 3 to the right side:
$t^2 - 6t = -3$

Step3: Complete the square

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

Step4: Rewrite as squared form

Left side is a perfect square, right side simplifies:
$(t - 3)^2 = 6$

Step5: Scale back to original coefficient

Multiply both sides by 16 to match original equation's leading coefficient:
$16(t - 3)^2 = 96$
Rearrange to match the required form:
$16(t - 3)^2 - 96 = 0$

Answer:

$16(t - 3)^2 = 96$ (or equivalent form $16(t - 3)^2 - 96 = 0$)