# What is the kinetic energy of an object with a mass of 1 kg that has been in freefall for 2 s?

Mar 9, 2016

I found $192 J$

#### Explanation:

We can either:

1) use kinematic to find the final velocity after $2 s$:
using: ${v}_{f} = {v}_{i} + a t$
where ${v}_{i}$ is the initial velocity (assumed as zero);
$a$ is the acceleration of gravity (directed downward).
We get:
${v}_{f} = 0 - 9.8 \cdot 2 = 19.6 \frac{m}{s}$
From this we get:
Kinetic Energy: $K = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot 1 \cdot {\left(19.6\right)}^{2} = 192 J$

or:

2) use kinematic (and Conservation of Energy) to find the vertical distance described after $2 s$ and the Potential Energy:
using: ${y}_{f} - {y}_{i} = {v}_{i} t + \frac{1}{2} a {t}^{2}$
Where we set ${y}_{f} = 0$ as the ground (zero Potential Energy).
giving:
$0 - {y}_{i} = 0 \cdot 2 - \frac{1}{2} \cdot 9.8 \cdot {\left(2\right)}^{2}$
so that the height (from ground) will be:
${y}_{i} = 19.6 m$
At this position the object had a Potential Energy of:
$U = m g {y}_{i} = 1 \cdot 9.8 \cdot 19.6 = 192 J$ that will be converted entirely into Kinetic Energy during the fall!