# The Graph Power

Consider the free fall of a particle at position $A$ with height $H$. Given that $AB = BC$ and the time for the particle to travel from $B$ to $C$ is $2\text{ s}$.

1. What is the time traveled from $A$ to $B$, in seconds?
2. What is the height $H$, in meters?

First, we draw a $v-t$ graph of the particle

We knew that $AB=BC$. So, the area of $A$ and $B$ must be the same, hence the equation

\begin{aligned} \displaystyle \frac{1}{2}\times (9.8x)(x) &= \frac{1}{2}\times 2[9.8x+9.8(x+2)] \\ x^2 &= 2(x+x+2) \\ x^2 - 4x - 4 &= 0 \\ x &= 4.828 \text{ s} \end{aligned}

The time traveled from $A$ to $B$ is $4.828 \text{ s}$.

$H$ is the area of the graph

\begin{aligned} \displaystyle \text{Area} &= \frac{1}{2}[9.8(4.828+2)](4.828+2) \\ &= 228.4\text{ m} \end{aligned}

Note:
In my opinion, it would be more complicated if we used elementary formula to solve this problem. Worse, they are easier to make mistakes and harder to look for errors. If one can use the graph wisely, we can optimize the problem and tackle it effectively.