Question

what is the total time the rock remains in the air before hitting the ground?
d = v_{i}t+\frac{1}{2}at^{2}
v_{i}=8.5 m/s
d = 100 m

Explanation:

Step1: Identify the kinematic - equation values

We are given the kinematic equation $d = V_{i}t+\frac{1}{2}at^{2}$, where $V_{i}=8.5\ m/s$, $d = 100\ m$, and assuming free - fall near the Earth's surface, $a = 9.8\ m/s^{2}$. Substituting these values, we get $100=8.5t + 4.9t^{2}$.

Step2: Rearrange the equation to quadratic form

Rearrange $100=8.5t + 4.9t^{2}$ to $4.9t^{2}+8.5t - 100 = 0$.

Step3: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 4.9$, $b = 8.5$, and $c=-100$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(8.5)^{2}-4\times4.9\times(-100)=72.25 + 1960=2032.25$.
Then, $t=\frac{-8.5\pm\sqrt{2032.25}}{2\times4.9}=\frac{-8.5\pm45.08}{9.8}$.
We have two solutions for $t$:
$t_1=\frac{-8.5 + 45.08}{9.8}=\frac{36.58}{9.8}\approx3.73\ s$
$t_2=\frac{-8.5 - 45.08}{9.8}=\frac{-53.58}{9.8}\approx - 5.47\ s$.
Since time cannot be negative in this context, we discard the negative solution.

Answer:

$t\approx3.73\ s$