Question

a ball is thrown directly downward with an initial speed of 8.45 m/s, from a height of 31.0 m. after what time interval does it strike the ground?

Explanation:

Step1: Identify the kinematic - equation

We use the equation $y = y_0+v_0t+\frac{1}{2}at^2$. Here, $y = 0$ (ground - level), $y_0=31.0$ m (initial height), $v_0 = 8.45$ m/s (initial velocity, downward), and $a = g=9.8$ m/s² (acceleration due to gravity, downward). So the equation becomes $0 = 31.0+8.45t - 4.9t^2$.

Step2: Rearrange to quadratic - form

The quadratic equation is $4.9t^2-8.45t - 31.0 = 0$. The quadratic formula for a quadratic equation $ax^2+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$. Here, $a = 4.9$, $b=-8.45$, and $c=-31.0$.

Step3: Calculate the discriminant

First, calculate the discriminant $\Delta=b^2 - 4ac=(-8.45)^2-4\times4.9\times(-31.0)=71.4025 + 607.6=679.0025$.

Step4: Find the time using the quadratic formula

$t=\frac{8.45\pm\sqrt{679.0025}}{9.8}=\frac{8.45\pm26.0577}{9.8}$. We have two solutions for $t$: $t_1=\frac{8.45 + 26.0577}{9.8}=\frac{34.5077}{9.8}\approx3.52$ s and $t_2=\frac{8.45-26.0577}{9.8}=\frac{-17.6077}{9.8}$ (we discard the negative solution since time cannot be negative in this context).

Answer:

$t\approx3.52$ s