Free fall

As you know, all objects released from a certain height fall on the ground. This, because there is a force (gravitational force) acting from the Earth to the objects near its surface that pulls them towards the center of the Earth. Therefore, these objects fall vertically down. In fact, free fall is a sub-category of the vertical motion (because vertical motion includes not only free fall but also situations where the objects are thrown upwards or when they are thrown downwards by giving them an initial velocity).

                                             

Since there is a vertical force acting on all objects near the Earth's surface, there is also an acceleration caused by the gravity. This acceleration is more or less the same on different places on the Earth's surface (because Earth is a sphere and the distance from the ground to its center is constant). Therefore, free fall is considered as uniformly accelerated motion. Therefore, all equations used to investigate the uniformly accelerated motion are also valid for the free fall. Furthermore, the magnitude of this acceleration has a fixed value (+9.81 m/s2 for all objects on the Earth surface, although we will round it up to 10 m/s2 in our examples simply for practical purposes) because the greatest part of the human activity takes place near the Earth's surface.

This acceleration caused by the gravity (also known as gravitational acceleration) is a vector quantity, as all the other types of acceleration. It is denoted by the symbol \((\vec{g})\) and is measured in m/s2. As you may remember, the equations of the uniformly accelerated motion are:

1-     \(\vec{v} = \vec{v_0}+\vec{a}\times{t}\)

2-   \(\vec{\Delta{x}}=\vec{v_0}\times{t}+\frac{\vec{a}\times{t}^2} {2}\)  

3-  \(\vec{v}^2-\vec{v_0}^2=2\times{\vec{a}}\times{\vec{\Delta{x}}} \)

Here, we have to make some small arrangements in order to obtain the equations of free fall. First, this motion is performed by starting from rest as the object are released and not thrown down. Therefore, the magnitude of the initial velocity\((\vec{v_0})\) is zero. Another element to be changed is the letter representing the vertical displacement (height) which is not written anymore by using the symbol \(\vec{\Delta{x}}\)  but \((\vec{h})\), (a short notation for height). Also, as mentioned before, the acceleration is written by using the letter \((\vec{g})\) instead of \((\vec{a})\). Thus, we obtain  3 new equations for the free fall:

1. \(\vec{v}=\vec{g}\times{t}\)

2. \(\vec{h}=\frac{\vec{g}\times{t}^2}{2}\)

3. \(\vec{v^2}=2\times{\vec{g}}\times{\vec{h}}\)


Note: In all situations regarding the free fall studied in this topic, we will neglect the effect of air resistance. The abovementioned equations are valid only if we ignore the air resistance.


Example 1: An object falls from 80 m above the ground. Find:

a) The time necessary to reach the ground

b) The velocity just before striking the ground

(take \(\vec{g}=10m/s^2\))

Solution:

a) The object falls without initial velocity. Therefore, 

\(\vec{h}=\frac{\vec{g}\times{t}^2}{2}=\frac{{10}\;m/s^2\times{t}^2}{2}=80\;m\)

Thus, \(10\;m/s^2\times{t}^2=2\times{80}\;m=160\;m\)

\(t^2=160\;m\div{10}\;m/s^2=16\;s^2\)

\(t=\sqrt{16\;s^2}=4\;s\)

b) The final velocity (the velocity just before striking the ground can be found in two ways: 

1. Using the first equation of free fall. Thus, 

\(\vec{v}=\vec{g}\times{t}=10\;m/s^2\times 4\;s=40\;m/s\)

2- Using the third equation of free fall. Thus, 

\(\vec{v}^2=2\times{\vec{g}\times{\vec{h}}}=2\times{10}\;m/s^2\times80\;m=1600\;m^2/s^2\)

\(\vec{v}=\sqrt{1600\;m^2/s^2}=40\;m/s\)

As you can see, the result obtained is the same in both methods used.


Example 2: An object is released from 320 m above the ground. What is the vertical displacement of the object during the last second of motion? (Take \(\vec{g}=10\;m/s^2\))

Solution: First we find the total time of motion by using the second equation of free fall (remember that during free fall the initial velocity is zero). Thus,

\(\vec{h}=\frac{\vec{g}\times{t}^2}{2}\)

\(2\times{\vec{h}}=\vec{g}\times{t}^2\)

\(t^2=\frac{2\times{\vec{h}}}{\vec{g}}\)

\(t= \frac{\sqrt{2\times{h}}}{\sqrt{g}} \)

\(t=\frac{\sqrt{2\times{320}}}{\sqrt{10}}\)

\(t=\sqrt{\frac{640}{10}}=\sqrt{64}=8\;s\)

The vertical displacement of the object during the last second can be easier calculated indirectly. Thus, first find the vertical displacement during the first \(8-1=7\;s\) and then, from the total height subtract the vertical displacement during the first 7 seconds. Hence, for the time \(t_1=7\;s\) we have:

\(\vec{h_1}=\frac{\vec{g}\times{t_1}^2}{2}=\frac{10\times{7^2}}{2}=5\times{49}\;m=245\;m.\)

Thus, the vertical displacement \(\begin{pmatrix} \vec{h_2} \end{pmatrix}\) during the last second will be:

\(\vec{h_2}=\vec{h} - \vec{h_1}=320\;m-245\;m=75\;m\)

Important! The value of gravitational acceleration changes slightly from place to place over the Earth's surface. For example, in equator it is 9.78 m/sand at poles it is 9.83 m/s2. This because the Earth is not a perfect sphere. The distance from the poles to the center of the Earth is greater than the distance from the equator to its center. Therefore, the effect of gravity is greater at poles than at equator. As an average value we use the value 9.81 m/s2.