Question

p40. a ball is thrown directly downward with an initial speed of 8.00 m/s from a height of 30.0 m. after what time interval does it strike the ground?

Explanation:

Step1: List known values

Knowns: $y_0 = 30.0\ \text{m}$, $y = 0\ \text{m}$, $v_0 = 8.00\ \text{m/s}$, $g = 9.81\ \text{m/s}^2$

Step2: Use kinematic equation

$$y = y_0 + v_0 t - \frac{1}{2} g t^2$$

Step3: Substitute values into equation

$$0 = 30.0 + 8.00t - \frac{1}{2}(9.81)t^2$$
Simplify to standard quadratic form:
$$4.905t^2 - 8.00t - 30.0 = 0$$

Step4: Solve quadratic equation

Use quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=4.905$, $b=-8.00$, $c=-30.0$:
$$t = \frac{8.00 \pm \sqrt{(-8.00)^2 - 4(4.905)(-30.0)}}{2(4.905)}$$
Calculate discriminant:
$$\sqrt{64 + 588.6} = \sqrt{652.6} \approx 25.55$$
Take positive root (time can't be negative):
$$t = \frac{8.00 + 25.55}{9.81} \approx \frac{33.55}{9.81}$$

Step5: Compute final time value

$$t \approx 3.42\ \text{s}$$

Answer:

$\boldsymbol{3.42\ \text{seconds}}$