Question

the height of an arrow shot upward can be given by the formula $s = v_0t - 16t^2$, where $v_0$ is velocity and $t$ is time. how long does it take for the arrow to reach a height of 48 ft if it has a 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 $(t - 3)^2 = 6$.
complete
solve $(t - 3)^2 = 6$. the arrow is at a height of 48 ft after approximately \boxed{} s and after \boxed{} s.
done

Explanation:

Step1: Take square roots of both sides

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

Step2: Simplify radicals

$$t-3 = \pm\sqrt{6}$$

Step3: Isolate t

$$t = 3 \pm \sqrt{6}$$

Step4: Calculate decimal values

$t_1 = 3 + \sqrt{6} \approx 3 + 2.45 = 5.45$
$t_2 = 3 - \sqrt{6} \approx 3 - 2.45 = 0.55$

Answer:

0.55 s and 5.45 s