# Question #8b7fa

Sep 6, 2015

The object is at a height of $\frac{3}{4} \cdot h$ from ground level.

#### Explanation:

So, you know that your object is being released from a height $h$ and that it reaches the ground in a time $t$.

This means that you can write

$h = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{= 0}} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

Now, let's assume that we can stop this object's motion at $\frac{t}{2}$ and see how far it fell from its initial position $h$. This distance, let's say ${h}_{1}$, will be

${h}_{1} = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{= 0}} \cdot \frac{t}{2} + \frac{1}{2} \cdot g \cdot {\left(\frac{t}{2}\right)}^{2}$

${h}_{1} = \frac{1}{2} \cdot g \cdot {t}^{2} / 4$

Notice that ${h}_{1}$ is actually equal to

${h}_{1} = {\underbrace{\frac{1}{2} \cdot g \cdot {t}^{2}}}_{\textcolor{b l u e}{= h}} \cdot \frac{1}{4} = \frac{1}{4} \cdot h$

Well, if the object travelled $\frac{h}{4}$ in $\frac{t}{2}$, then its distance from the ground is

${h}_{\text{from ground}} = h - {h}_{1}$

${h}_{\text{from ground}} = h - h \cdot \frac{1}{4} = \textcolor{g r e e n}{\frac{3}{4} \cdot h}$