A particle moving in a parabolic path in the ##x-y## plane

[brotherbobby]

Then what else do you need to find velocity at time t=0? What is the $v_0$ in the expression for q?

I copy and paste my working below.

This is the answer from the book, but it makes reference to the details of projectile motion, so perhaps less suitable. I suspect the way to the answer is @Orodruin 's method in post#5, which I will do in a moment.
As you see in the posts above, I am stuck with a conceptual problem - namely, should the equation of the trajectory of a projectile change if one were to effect a change in its coordinates to some point along the trajectory?

 

[PeroK]

As you see in the posts above, I am stuck with a conceptual problem - namely, should the equation of the trajectory of a projectile change if one were to effect a change in its coordinates to some point along the trajectory?

Imagine you fire a projectile straight up and want the speed at the origin, which was the original starting point. First, you move the origin to the highest point, where the speed is zero and that's your answer?

 

[brotherbobby]

Imagine you fire a projectile straight up and want the speed at the origin, which was the original starting point. First, you move the origin to the highest point, where the speed is zero and that's your answer?

I am sure you phrased the question incorrectly. Can you rephrase it for me?

 

[PeroK]

I am sure you phrased the question incorrectly. Can you rephrase it for me?

I think you should focus on the problem you started with. As a separate exercise you might want to study some graphs and look at how their equations change when you translate the $x$ and $y$ coordinates.

If you don't like taking $t = 0$ at the origin, then take $t = t_0$ at the origin. Or, use the alternative method using derivatives, where you don't need to introduce the time parameter.

In any case, you don't need the parameter $\theta$. That unnecessarily complicates the working. You are already given $p$ and $q$. That's what you should be working with.

 

[brotherbobby]

You already know that $x = v_x t$. This means that any derivatives with respect to $t$ may be rewritten as derivatives with respect to $x$ as $df/dt = v_x (df/dx)$. For the second derivative you would have $d^2f/dt^2 = v_x^2 (d^2f/dx^2)$. Use these relations for $f = y$ and evaluate at the origin. Poff, two unknowns ($v_x$ and $v_y$) and two equations.

Can you check the last line of your statement @Orodruin ?

This is what I am getting.

I have an added term for equation $2$. I wonder if I am mistaken.

 

[PeroK]

Introducing the function $f$ is a red herring

To get you started, here's what I would do (note that $y = px - qx^2$):
$$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} = (p - 2qx)v_x$$Where $v_x$ is a constant.

And, here's another big hint. In addition to the constant $v_x$, we need to find $v_y(0) \equiv v_y(x = 0)$.

 

[Orodruin]

Introducing the function f is a red herring

I disagree. It is just showing the general rule and then using f = y.

I have an added term for equation $2$. I wonder if I am mistaken.

No, but $v_x$ is constant so I omitted that term.

 

[Orodruin]

Another note, the last term can also be rewritten:
$$
v_x \frac{dv_x}{dx} f’(x) = \frac{dx}{dt} \frac{dv_x}{dx} f’(x)
= \frac{dv_x}{dt} f’(x) = a_x f’(x)
$$
As per the problem statement $a_x = 0$.

 

[PeroK]

This is more advanced than using the equations of projectile motion - which is the book solution. Also, at this level, I would use the given function to keep the complications to a minimum.

 

[brotherbobby]

Yes, I got it. Let me copy and paste the solution to the problem. Better still, since we have had plenty to discusss and digressions along the way, let me start with the problem itself.

Problem statement :Solution : Given the equation of the curve : $y = px-qx^2$. Thus
\begin{equation*}
\begin{split}
\boldsymbol{\dot y}& = \boldsymbol{p\dot x-2qx\dot x}\\
\Rightarrow \dot y(O)& = p\dot x(O)\quad [x(O) = 0]\\
\Rightarrow v_y(O)&=pv_x(O)\quad\quad \mathbf{(1)}
\end{split}
\end{equation*}
Differentiating the bold equation above again, we get
\begin{equation*}
\begin{split}
\ddot y& = \cancel{p\ddot x}-2q\dot x^2-\cancel{2qx\ddot x}\quad [a_x=0\rightarrow \ddot x=0, x(O)=0]\\
\Rightarrow \ddot y(O)& = -2q\dot x(O)^2\\
\end{split}
\end{equation*}
But it is given that $\ddot y = -a_0$, since the acceleration is directed downward. Hence from above, we obtain $-a_0 = -2q\dot x(O)^2\Rightarrow v_x(O) = \sqrt{\dfrac{a_0}{2q}}\quad\quad \mathbf{(2)}$

Using (1) above (2) above, we have $v_y(O) = \sqrt{\dfrac{a_0}{2q}} p$

But $v(O) = \sqrt{v_x(O)^2+v_y(O)^2} \Rightarrow \boxed{\boldsymbol{v(O) = \sqrt{\dfrac{a_0}{2q}(1+p^2)}}}\quad\color{green}{\Huge{\checkmark}}$.

This agrees with the answer in the textbook.{The problem fades in importance to the critical point I gathered during my discussions with @PeroK - namely, that the equation of the path of a projectile $y(x) = px-qx^2$ is true only for the initial velocity $v_0$ defined at the origin. I am yet to show how this is true, and the thread therefore remains unresolved}

 

[PeroK]

namely, that the equation of the path of a projectile $y(x) = px-qx^2$ is true only for the initial velocity $v_0$ defined at the origin. I am yet to show how this is true, and the thread therefore remains unresolved}

You can have $v_0$ wherever you like, but then it's no longer $v_0$ that you are trying to calculate! If you are asked for the velocity at the spatial origin, it makes little sense to set $v_0$ somewhere else.

Moreover, if you don't have $t = 0$ at the origin, then you need to be careful about whether $v_0$ is $v(t = 0)$ or $v(x = 0)$ as these no longer coincide.