Question

a swimmer bounces straight up from a diving board and falls feet first into a pool (ignore any horizontal motion). she starts with a velocity an upward velocity of 4.00 m/s and her takeoff point is 8 m above the pool. how long are her feet in the air?
if your answer is not an integer, write it to 3 significant figures in standard decimal form.
______ s
your answer

Explanation:

Step1: Define kinematic equation

$$\Delta y = v_0 t + \frac{1}{2} a t^2$$

Step2: Assign known values

Let $\Delta y = -8\ \text{m}$ (displacement downward), $v_0 = 4.00\ \text{m/s}$, $a = -9.81\ \text{m/s}^2$. Substitute:
$$-8 = 4.00t - \frac{1}{2}(9.81)t^2$$
Rearrange to standard quadratic form:
$$4.905t^2 - 4.00t - 8 = 0$$

Step3: Solve quadratic equation

Use quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=4.905$, $b=-4.00$, $c=-8$:
$$t = \frac{4.00 \pm \sqrt{(-4.00)^2 - 4(4.905)(-8)}}{2(4.905)}$$
Calculate discriminant:
$$\sqrt{16 + 156.96} = \sqrt{172.96} \approx 13.15$$
Take positive root (time can't be negative):
$$t = \frac{4.00 + 13.15}{9.81} \approx \frac{17.15}{9.81}$$

Step4: Compute final value

$$t \approx 1.748$$
Round to 3 significant figures.

Answer:

1.75 s