I Tracing parabolic motion with only current velocity and position?

AI Thread Summary
Tracing the trajectory of an object using only its velocity and position is feasible by applying the equations of motion under constant gravitational force. The initial conditions, including initial velocity and position, are critical for accurate calculations. The method involves integrating the equations of motion to derive the object's position over time, accounting for gravitational acceleration. However, the challenge arises when the vertical velocity decreases due to gravity, affecting the perceived maximum height. Ultimately, solving the equations of motion provides a comprehensive understanding of the object's trajectory.
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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.

$$time_{max} = \frac{v_{y}}{g}\space \space(1)$$
$$x_{max} = x_{current} + v_{x} * time_{max}\space \space(2)$$
$$y_{max} = y_{current} + \frac{{v_{y}}^2}{2a}\space \space(3)$$
$$parabola: a(x-x_{max})^2 + y_{max}\space \space(4)$$
 
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question_asker said:
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.
 
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.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
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