# 0.3 Gravity and mechanical energy  (Page 4/9)

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A ball is dropped from the balcony of a tall building. The balcony is $15\phantom{\rule{2pt}{0ex}}m$ above the ground. Assuming gravitational acceleration is $9,8\phantom{\rule{2pt}{0ex}}m·s{}^{-2}$ , find:

1. the time required for the ball to hit the ground, and
2. the velocity with which it hits the ground.
1. It always helps to understand the problem if we draw a picture like the one below:

2. We have these quantities:

$\begin{array}{ccc}\hfill \Delta x& =& 15\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \\ \hfill {v}_{i}& =& 0\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \\ \hfill g& =& 9,8\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-2}\hfill \end{array}$
3. Since the ball is falling, we choose down as positive. This means that the values for ${v}_{i}$ , $\Delta x$ and $a$ will be positive.

4. We can use [link] to find the time: $\Delta x={v}_{i}t+\frac{1}{2}g{t}^{2}$

5. $\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}g{t}^{2}\hfill \\ \hfill 15& =& \left(0\right)t+\frac{1}{2}\left(9,8\right){\left(t\right)}^{2}\hfill \\ \hfill 15& =& 4,9\phantom{\rule{3.33333pt}{0ex}}{t}^{2}\hfill \\ \hfill {t}^{2}& =& 3,0612...\hfill \\ \hfill t& =& 1,7496...\hfill \\ \hfill t& =& 1,75\phantom{\rule{3.33333pt}{0ex}}s\hfill \end{array}$
6. Using [link] to find ${v}_{f}$ :

$\begin{array}{ccc}\hfill {v}_{f}& =& {v}_{i}+gt\hfill \\ \hfill {v}_{f}& =& 0+\left(9,8\right)\left(1,7496...\right)\hfill \\ \hfill {v}_{f}& =& 17,1464...\hfill \end{array}$

Remember to add the direction: ${v}_{f}=17,15\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ downwards.

By now you should have seen that free fall motion is just a special case of motion with constant acceleration, and we use the same equations as before. The only difference is that the value for the acceleration, $a$ , is always equal to the value of gravitational acceleration, $g$ . In the equations of motion we can replace $a$ with $g$ .

## Gravitational acceleration

1. A brick falls from the top of a $5\phantom{\rule{2pt}{0ex}}m$ high building. Calculate the velocity with which the brick reaches the ground. How long does it take the brick to reach the ground?
2. A stone is dropped from a window. It takes the stone $1,5\phantom{\rule{2pt}{0ex}}s$ to reach the ground. How high above the ground is the window?
3. An apple falls from a tree from a height of $1,8\phantom{\rule{2pt}{0ex}}m$ . What is the velocity of the apple when it reaches the ground?

## Potential energy

The potential energy of an object is generally defined as the energy an object has because of its position relative to other objects that it interacts with. There are different kinds of potential energy such as gravitional potential energy, chemical potential energy, electrical potential energy, to name a few. In this section we will be looking at gravitational potential energy.

Potential energy

Potential energy is the energy an object has due to its position or state.

Gravitational potential energy is the energy of an object due to its position above the surface of the Earth. The symbol $PE$ is used to refer to gravitational potential energy. You will often find that the words potential energy are used where gravitational potential energy is meant. We can define potential energy (or gravitational potential energy, if you like) as:

$PE=mgh$

where PE = potential energy measured in joules (J)

m = mass of the object (measured in kg)

g = gravitational acceleration ( $9,8\phantom{\rule{2pt}{0ex}}m·s{}^{-2}$ )

h = perpendicular height from the reference point (measured in m)

A suitcase, with a mass of $1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ , is placed at the top of a $2\phantom{\rule{2pt}{0ex}}m$ high cupboard. By lifting the suitcase against the force of gravity, we give the suitcase potential energy. This potential energy can be calculated using [link] .

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