Parabolic movement derivations
On this page, the formulas for the maximum height \( h_{\mathrm{max}} \), the range \( R \), and the flight time \(T\) are derived for cases without an initial height \( y_{0} \) and with it. Remember that: \( \vec{a} = \mathbf{a} \), \(\dot{a}= \frac{da}{dt}\) and \(\Delta a = a_{final} - a_{initial} = a_{f} - a_{i} = a_{2} - a_{1} \). The \(\propto\) symbol is read as "is proportional to".
Start in origin \(y_{0} = 0\)
Assuming the projectile is launched from the origin \((x_0, y_0) = (0,0)\), the trajectory equations for the horizontal and vertical components are:
$$x(t)=v_{0}t\cos{\theta},$$
$$y(t)=v_{0}t\sin{\theta}-\frac{1}{2}gt^{2}.$$
To simplify the equations, we express the initial velocity \( v_0 \) in terms of its components:
$$v_{0_{x}}=v_{0}\cos{\theta}$$
$$v_{0_{y}}=v_{0}\sin\theta$$
Substituting these expressions into the trajectory equations:
$$x(t)=v_{0_{x}}t$$
$$y(t)=v_{0_{y}}t-\frac{1}{2}gt^{2}$$
Using the velocity equation \( v = x / t \), we solve for time:
$$t=\frac{x}{v_{0_{x}}}$$
Substituting this into the vertical trajectory equation:
$$\begin{split} y &= v_{0_{y}}t - \frac{1}{2}gt^{2} \\ &=v_{0_{y}}\left(\frac{x}{v_{0_{x}}}\right) - \frac{1}{2}g\left(\frac{x}{v_{0_{x}}}\right)^{2} \\ &= \left(\frac{-g}{2v^{2}_{0_{x}}}\right)x^{2} + \left(\frac{v_{0_{y}}}{v_{0_{x}}}\right)x \end{split}$$
Maximum height \(h_{\mathrm{max}}\)
The maximum height can be obtained by differentiating the trajectory equation.
$$\begin{split}\frac{dy}{dx} &= 2\left(\frac{-g}{2v^{2}_{0_{x}}}\right)x + \left(\frac{v_{0_{y}}}{v_{0_{x}}}\right) \\ &= - \left(\frac{g}{v^{2}_{0_{x}}}\right)x + \left(\frac{v_{0_{y}}}{v_{0_{x}}}\right)\end{split}$$
Setting this expression equal to zero allows us to find the critical value of \(x\).
$$ \begin{split}0 &= - \left(\frac{g}{v^{2}_{0_{x}}}\right)x + \left(\frac{v_{0_{y}}}{v_{0_{x}}}\right) \\ \Rightarrow x &= \left(\frac{v_{0_{y}}v^{2}_{0_{x}}}{gv_{0_{x}}}\right) = \frac{v_{0_{x}}v_{0_{y}}}{g}\end{split}$$
Applying the second derivative test:
$$\frac{d^{2}y}{dx^{2}} = - \left(\frac{g}{v^{2}_{0_{x}}}\right)$$
Since this second derivative is a negative constant, the critical value corresponds to a maximum. Substituting this value into the trajectory equation, we obtain the maximum height:
$$\begin{split}y &= \left(\frac{-g}{2v^{2}_{0_{x}}}\right) \left(\frac{v_{0_{x}}v_{0_{y}}}{g}\right)^{2} + \left(\frac{v_{0_{y}}}{v_{0_{x}}}\right) \left(\frac{v_{0_{x}}v_{0_{y}}}{g}\right) \\ &= -\frac{v^{2}_{0_{y}}}{2g} + \frac{v^{2}_{0_{y}}}{g} \\ &= \frac{-v^{2}_{0_{y}} + 2v^{2}_{0_{y}}}{2g} \\ &= \frac{v^{2}_{0_{y}}}{2g}\end{split}$$
Maximum Range \(R\)
In the trajectory equation, we consider the case when \(y = 0\).
$$0 = \left(\frac{-g}{2v^{2}_{0_{x}}}\right)x^{2} + \left(\frac{v_{0_{y}}}{v_{0_{x}}}\right)x$$
Factoring \(x\), we obtain:
$$0 = x\left[\left(\frac{-g}{2v^{2}_{0_{x}}}\right)x + \left(\frac{v_{0_{y}}}{v_{0_{x}}}\right)\right]$$
For this equation to hold, \(x\) must be equal to zero or:
$$\begin{split}x &= \frac{-\frac{v_{0_{y}}}{v_{0_{x}}}}{-\frac{g}{2 v^{2}_{0_{x}}}} \\ &= \frac{2v^{2}_{0_{x}}v_{0_{y}}}{gv_{0_{x}}} \\ &= \frac{2v_{0_{x}}v_{0_{y}}}{g} \end{split}$$
This is the maximum range, represented by the letter \(R\), which comes from the English term Range.
$$R = \frac{2v_{0_{x}}v_{0_{y}}}{g}$$
Flight Time \(T\)
We use the velocity equation \(v_{0_x} = x/t\) to determine the flight time based on the range.
$$\begin{split}T &= \frac{R}{v_{0_{x}}} \\ &= \frac{\frac{2v_{0_{x}}v_{0_{y}}}{g}}{v_{0_{x}}} \\ &= \frac{2v_{0_{y}}}{g}\end{split}$$
Start out of origin \(y_{0} \neq 0\)
The following trajectory equation is used: \( y = y_{0} + v_{0_y}t - \frac{1}{2}gt^{2} \), which describes motion when the initial height is different from zero.
Maximum Height \(h_{\mathrm{max}}\)
We rewrite the trajectory equation in its canonical form by completing the square.
$$\begin{split}y - {y_{0}} &= - \frac{1}{2}g\left(t^{2} - \frac{2v_{0_{y}}t}{g}\right) \\ &= - \frac{1}{2}g\left[t^{2} - 2\left(\frac{v_{0_{y}}}{g}\right)t + \frac{v^{2}_{0_{y}}}{g^{2}}\right] - \left( - \frac{1}{2}g\right)\left(\frac{v^{2}_{0_{y}}}{g^{2}}\right) \end{split}$$
Thus:
$$\begin{split}&y - y_{0} = -\frac{1}{2}g\left(t - \frac{v_{0_{y}}}{g}\right)^{2} + \frac{v^{2}_{0_{y}}}{2g} \\ \Rightarrow &y - y_{0} - \frac{v^{2}_{0_{y}}}{2g} = - \frac{1}{2}g\left(t - \frac{v_{0_{y}}}{g}\right)^{2}\end{split}$$
$$\begin{split}\therefore \left( t - \frac{v_{0_{y}}}{g} \right)^{2} &= - \frac{2}{g}\left[y - \left(y_{0} + \frac{v^{2}_{0_{y}}}{2g}\right)\right] \\ \equiv (x - h)^{2} &= 4p(y - k)^{2}\end{split}$$
Therefore, the vertex of the parabola is:
$$t = \frac{v_{0_{y}}}{g}$$
$$h_{\max} = y_{0} + \frac{v^{2}_{0_{y}}}{2g}$$
This can also be obtained by differentiating the trajectory equation \(y(t) = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}\) and setting it to zero.
$$\begin{split} y'(t) = v_{0_{y}} - gt &= 0 \\ \Rightarrow v_{0_{y}} &= gt \\ \Rightarrow t &= \frac{v_{0_{y}}}{g} \end{split}$$
Substituting into the trajectory equation:
$$\begin{split}\Rightarrow y(t) &=y_{0}+v_{0_{y}}\left( \frac{v_{0_{y}}}{g}\right)-\frac{1}{2}g\left(\frac{v_{0_{y}}}{g}\right)^{2} \\ &= y_{0}+\frac{v_{0_{y}}^{2}}{g}-\frac{1}{2}\frac{v_{0_{y}}^{2}}{g} \\ &=y_{0}+\frac{v_{0_{y}}^{2}}{2g} \end{split}$$
$$h_{\max} = y_{0} + \frac{v^{2}_{0_{y}}}{2g}$$
Tiempo de vuelo \(T\)
To determine the flight time, we set the trajectory equation to zero:
$$y = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$$
Solving the equation with the following coefficients:
$$\begin{split} a &= - \frac{1}{2}g \\ b &= v_{0_{y}} \\ c &= y_{0} \end{split}$$
Substituting these values into the quadratic equation:
$$\begin{split} T &= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \\ T &= \frac{-v_{0_{y}}\pm \sqrt{v^{2}_{0_{y}} - 4\left(- \frac{1}{2}g\right)\left(y_{0}\right)}}{2\left(-\frac{1}{2}g\right)} \\ T &= \frac{v_{0_{y}}\pm \sqrt{v^{2}_{0_{y}}+2gy_{0}}}{g} \end{split}$$
Maximum Range \(R\)
The maximum range is reached when the flight time ends:
$$R = v_{0_{x}}T$$