# Question 79d47

Mar 4, 2017

$\text{21.3 m}$

#### Explanation:

The first thing to do here is to figure out the time needed for the rock to hit the ground.

Your tool of choice here is the equation

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

Here

• $h$ is the distance covered by the object in a given time $t$
• ${v}_{0}$ is the initial velocity of the rock
• $t$ is the time of flight
• $g$ is the gravitational acceleration, usually given as ${\text{9.81 m s}}^{- 2}$

In your case, the object starts from rest, so

${v}_{0} = {\text{0 m s}}^{- 1}$

This means that you will have

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

Rearrange to solve for $t$

$t = \sqrt{\frac{2 \cdot h}{g}}$

Plug in your values to find

t = sqrt( (2 * 28.5 color(red)(cancel(color(black)("m"))))/(9.81color(red)(cancel(color(black)("m")))"s"^(-2))) = "2.41 s"

So, you know that it takes $\text{2.41 s}$ for the rock to hit the ground. Use this value to determine the time spent falling before reaching the $\text{1.2-s}$ mark

${t}_{\text{needed" = "2.41 s" - "1.2 s" = "1.21 s}}$

Use this value in the above equation and solve for $h$, the distance covered by the rock in the first $\text{1.21 s}$ of falling

h_"1.21 s" = 1/2 * 9.81 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2)))) * (1.21)^2 color(red)(cancel(color(black)("s"^2)))#

${h}_{\text{1.21 s" = "7.18 m}}$

Therefore, you can say that $\text{1.21 s}$ after it began to fall and $\text{1.2 s}$ before hitting the ground, the rock was

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{h}_{\text{above ground" = "28.5 m" - "7.18 m" = "21.3 m}}}}} \to$ above the ground