Tracing parabolic motion with only current velocity and position?

[question_asker]

TL;DR Summary

Using only the current component-wise velocity and position, the trajectory of a projectile needs to be found. The object may change velocity mid flight, an the trajectory should reflect any changes.

Is it possible to trace the trajectory of an object using only its velocity and position, both of which are given as components. My method of doing so involves using the time until max height is reached, and using that time value to calculate the max height itself (h,k), then plugging in the current point to find the constant a in the equation y=a(x-h^2) +k. I used the following equations, but noticed that as the vertical velocity decreased due to gravity, the max height also changes which it doesn't in reality. I am now wondering if what I'm trying to do is possible or not.

$$time_{max} = \frac{v_{y}}{g}\space \space(1)$$
$$x_{max} = x_{current} + v_{x} * time_{max}\space \space(2)$$
$$y_{max} = y_{current} + \frac{{v_{y}}^2}{2a}\space \space(3)$$
$$parabola: a(x-x_{max})^2 + y_{max}\space \space(4)$$

 

[A.T.]

I used the following equations, but noticed that as the vertical velocity decreased due to gravity, the max height also changes which it doesn't in reality.

You are using the remaining climb time to max height which does decrease, and so does the remaining height.

 

[vanhees71]

Obviously you think about the initial-value problem for motion under the influence of the contant gravitational force of the Earth on a point particle. Just solve the equations of motion, which is very simple:
$$m \ddot{\vec{x}}=m \vec{g} \; \Rightarrow \; \ddot{\vec{x}}=\vec{g}=\text{const}.$$
This is to be solved, assuming the initial condition: $\vec{v}(0)=\vec{v}_0$ and $\vec{x}(0)=\vec{x}_0$.

Integration the EoM once, using this initial condition gives
$$\int_0^{t} \mathrm{d} t' \ddot{\vec{x}}(t')=\dot{\vec{x}}(t)-\dot{\vec{x}}(0)=\dot{\vec{t}}-\vec{v}_0 = \int_0^t \mathrm{d} t' \vec{g}=g t.$$
So you get
$$\dot{\vec{x}}(t)=\vec{v}_0+\vec{g} t.$$
Integrating once more wrt. time in the analogous way, you finally get the solution
$$\vec{x}(t)=\vec{x}_0 + \vec{v}_0 t + \frac{1}{2} \vec{g} t^2.$$
Of course, now you can use this to figure out all kinds of different representations of this solution and analyze its properties.