Free Fall

The easiest example to explain the power of the equation of motion is the description of an object falling in the gravitational field of the earth. Experiments have shown that all objects near the surface of the earth fall with a constant acceleration which is called gravitational acceleration and is usually donated as \(g\). The value of \(g\) has been measured to \[g = 9.807\,\frac{\mathrm{m}}{\mathrm{s}^2}\] which can be seen as an average for different positions on earth. However, in most of the cases it is sufficient to approximate \(g\) with the value \(10\,\mathrm{m}/\mathrm{s}^2\). If an object falls down from the height \(h\) along the $y$-axis, its position with respect to time \(t\) can be derived from the general space-time curvature and written as follows: \[y(t) = -\frac{1}{2}gt^2 + h\] The minus sign denotes the movement of the object in the negative \(y\)-direction. If the initial height is known, the time until hitting the ground floor can be calculated by equating this formula with 0: $$-\frac{1}{2}gt^2 + h = 0$$ In the next step, the resulting equation has to be solved for the time \(t\) which leads to $$t=\pm\sqrt\frac{2h}{g}$$ The \(\pm\) signs follow from the calculation of the square root. However, the minus sign leads to unphysical results and can therefore be removed. The resulting formula is then given as: $$\boxed{t=\sqrt\frac{2h}{g}}$$ If one is interested in the final speed of the object just before hitting the ground, this formula can be inserted into the correlation between the speed of an object and the gravitational acceleration: \[v = gt = g\sqrt{\frac{2h}{g}}\] From that, the following result can be obtained: \[\boxed{v = \sqrt{2gh}}\]