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

[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.