## 2 Theoretical Background  
In the modern era, our understanding of gravity has evolved significantly, propelled by advancements in theoretical physics and observational evidence. At the heart of this understanding lies [PERSON]'s general theory of relativity (GR), which revolutionized our conception of gravity by proposing that massive objects warp the fabric of spacetime, causing curvature and influencing the motion of other objects. However, in this research, we focus on explaining the classical foundations of gravity as the main focus. We begin by revisiting [PERSON]'s law of universal gravitation, which states that the gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically, this can be expressed as  
\[F=G\frac{m_{1}m_{2}}{r^{2}} \tag{1}\]  
where \(F\) is the gravitational force between the two objects, \(G\) is the gravitational constant (\(6.674\times 10^{-11}\)m\({}^{3}\)kg\({}^{-1}\)s\({}^{-2}\)), \(m_{1}\) and \(m_{2}\) are the masses of the two objects, and \(r\) is the distance between the centers of the two objects. On the surface of the Earth, the object's mass (\(m_{2}\)) does not affect the value of \(g\) because it cancels out when considering the force experienced by an object of mass \(m_{2}\). Therefore, we usually consider \(m_{2}\) to be the mass of a test object experiencing gravity, and for simplicity, we can take \(m_{2}\) to be 1 kg. On the surface of the Earth, the distance (\(r\)) from the center of the Earth to the object's surface is approximately equal to the Earth's radius (\(R\)). Now, considering a test object with mass \(m_{2}=1\) kg, the gravitational force experienced by it near the surface of the Earth is  
\[F=G\frac{m_{1}m_{2}}{r^{2}}=G\frac{m_{1}}{R^{2}} \tag{2}\]  
Now, according to [PERSON]'s second law of motion, the force \(F\) acting on an object of mass \(m_{2}\) is equal to the mass times the acceleration (\(F=m_{2}g\)). So we have  
\[G\frac{m_{1}}{R^{2}}=m_{2}g \tag{3}\]  
\[g=G\frac{m_{1}}{R^{2}} \tag{4}\]  
Given that \(G=6.674\times 10^{-11}\) m\({}^{3}\)kg\({}^{-1}\)s\({}^{-2}\), \(R\approx 6.371\times 10^{6}\) m (radius of the Earth), and \(m_{1}\) is the mass of the Earth, which is approximately \(5.972\times 10^{24}\) kg, we can calculate \(g=9.81\) m/s\({}^{2}\). So the acceleration due to gravity on the surface of the Earth is approximately \(9.81\) m/s\({}^{2}\). However, when the elevation of the Earth's surface changes, the gravitational value (g) will also change.  
A solid body is permitted to fall freely over a fixed distance, denoted as \(H\), and the time taken to cover this distance is observed. The formula for a body, initially moving at a velocity of u m/s and traversing a distance \(H\) under free fall for a time t, is as follows:  
\[H=ut+\frac{1}{2}gt^{2} \tag{5}\]  
When a solid body is released from rest, with \(u\) = 0 m/s, Equation (5) can be rewritten as  
\[H=\frac{gt^{2}}{2} \tag{6}\]  
The calculation of \(g\) via the free fall method entails the precise measurement of the time (\(t\)) required for a steel ball to descend a defined vertical distance (\(H\)). The determination of \(g\) is derived through the application of the following mathematical relationship in Equation (6). In the context of this experiment, where \(H\) represents the falling height and \(t\) denotes the time of descent, the calculation of \(g\) is achieved by employing the following formula:  
\[g=\frac{2H}{t^{2}} \tag{7}\]