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 $\boldsymbol{(t - 3)^2 = 6}$
solve $(t - 3)^2 = 6$ the arrow is at a height of 48 ft after approximately \boxed{} s
and after \boxed{} s

Explanation:

Step1: Simplify the quadratic equation

Divide $16t^2 - 96t + 48 = 0$ by 16:
$$t^2 - 6t + 3 = 0$$

Step2: Rearrange for completing the square

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

Step3: Complete the square

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

Step4: Solve for t

Take square roots of both sides:
$$t - 3 = \pm\sqrt{6}$$
$$t = 3 \pm \sqrt{6}$$

Step5: Calculate numerical values

$\sqrt{6} \approx 2.449$, so:
$t_1 = 3 + 2.449 = 5.45$
$t_2 = 3 - 2.449 = 0.55$

Answer:

0.55 s
5.45 s