Relation between horizontal and vertical velocity in parabolic path So our teacher asked this question:

A particle P is travelling from A to B in a straight line with a velocity v.
Another particle Q is also travelling from A to B but in a parabolic path with a horizontal velocity v.
Which one takes lesser time?

He said Q takes lesser time and gave a logic that it's velocity increases till half its path and then decreases till it becomes v in the other half which I didn't quite get.
Can anyone explain this to me?
 A: If A is the origin and B is (8,6) in meters, for example.
Then if the parabola is designed to go through A and B and the horizontal component is 2m/s, it takes 4 seconds to reach B, (blue and green line below).
For the straight line path, the distance between A and B is 10 units.  At 2m/s it takes 5 seconds.
The parabola is quicker unless AB is horizontal, then the times are the same.

Hopefully this explains what your teacher probably meant.
A: The problem statement seems to say that

*

*the linear path from A to B has speed $V_{lin}=v$.

*the parabolic path from A to B has $V_{par,x}=v$
Expanding on @JohnHunter's answer...
Since the time of flight from A to B is
$$t_{par}=\frac{\Delta x}{V_{par,x}} \quad= \frac{\Delta x}{(\ v\ )},$$
let's get corresponding equation for the linear path
by computing $V_{lin,x}$, the "x-component of $\vec V_{lin}$".

The angle of $\vec V_{lin}$ is gotten by $$\tan V_{lin,\theta}=\frac{\Delta y}{\Delta x}.$$
(Of course, this angle is not equal to the angle of the launch velocity for the parabolic path.)
So,
$$\cos V_{lin,\theta}=\frac{\Delta x}{\sqrt{ (\Delta x)^2+(\Delta y)^2 }}$$
Thus,
\begin{align}
V_{lin,x}&=V_{lin}\cos V_{lin,\theta}\\
&=V_{lin}\frac{\Delta x}{\sqrt{ (\Delta x)^2+(\Delta y)^2 }}\\
&=(\ v\ )\frac{\Delta x}{\sqrt{ (\Delta x)^2+(\Delta y)^2 }}\\
\end{align}
as expected.

So now, putting things together:
$$t_{par}=\frac{\Delta x}{V_{par,x}} \quad= \frac{\Delta x}{(\ v\ )},$$
and now
$$t_{lin}=\frac{\Delta x}{V_{lin,x}} \quad= \frac{\Delta x}{ (\ v\ )\frac{\Delta x}{\sqrt{ (\Delta x)^2+(\Delta y)^2 }} }=\frac{ \sqrt{ (\Delta x)^2+(\Delta y)^2 } }{(\ v\ )}.$$
With given conditions placed on their speeds (from above)

*

*the linear path from A to B has speed $V_{lin}=v$

*the parabolic path from A to B has $V_{par,x}=v$,

the time of flight for the linear path is longer
than that for the parabolic path
due to the $\Delta y$ term.

...in accord with @JohnHunter's answer.
