I) I'm not sure if this is relevant for what OP is asking, but imagine a 1 dimensional motion with constant (negative) jerk $j<0$, see e.g. the second half of this webpage. Let $v_0>0$ be an unknown positive initial velocity, and $a_0$ be a given initial acceleration. We interpret OP's question(v1) as
For a given stopping distance $s>0$, what is the initial velocity $v_0$?
Well, the final velocity is zero. This yields an equation
$$\tag{1} 0~=~ v_f ~=~ v_0 + a_0 \Delta t + \frac{j}{2} \Delta t^2, $$
which is quadratic in time. There is one positive root
$$\tag{2} \Delta t ~=~ - \frac{\sqrt{a_0^2 -2j v_0}+a_0}{j} ~>~0. $$
The distance becomes
$$\tag{3}s ~=~ v_0 \Delta t+ \frac{a_0}{2} \Delta t^2+ \frac{j}{6} \Delta t^3 ~\stackrel{(2)}{=}~\frac{a_0(a_0^2-3jv_0)+ (a_0^2 -2j v_0)^{3/2}}{3j^2}.$$
For given $s>0$, $j<0$, and $a_0$, this equation (3) can be solved numerically for $v_0$. This ends our main answer.
II) As a consistency check of the above equations (2) and (3) in the case $a_0<0$, it is an instructive exercise to Taylor expand (or use l'Hopital's rule) in small $j$ to recover the well-known constant acceleration (suvat) formulas
$$\tag{2'} \Delta t ~\longrightarrow~ - \frac{v_0}{a_0}~>~0 \qquad \text{for} \qquad j~\longrightarrow ~0, $$
and
$$\tag{3'} s ~\longrightarrow~ - \frac{v^2_0}{2a_0}~>~0 \qquad \text{for} \qquad j~\longrightarrow ~0. $$
You do not have enough information. Time in the air, $t$, and maximum height, $h$, are both a function of the vertical launch velocity, $v_y$ only:
$$t = 2 \frac{v_y}{g}$$
$$h = \frac{v_y^2}{2g}$$
Your horizontal range requires knowing the horizontal velocity, which you cannot figure out from the data you have.
Best Answer
1- the maximum velocity is the maximum value of $\omega$ considered as a function of $x$. Just derive it twice with respect to $x$ and set the condition for a maximum (I will put my solution at the end, in case you want to try on your own)
2- the velocity profile is $\omega(x, t=t^*)$ for a fixed time $t^*$. It's just the value of the function $\omega$ at a given time, which is now just a function of $x$.
$\textbf{Solution}$:
1- $${d\omega \over dx}={\Lambda\over (4\nu t)^{3\over 2}}e^{-{x^2\over (4\nu t)}}\left(1-{x^2\over 2\nu t}\right) $$ which is equal to zero for $$x=\pm\sqrt{2\nu t}$$
Thus the max velocity is $$\omega(x=\sqrt{2\nu t}, t)={\Lambda\over 4\nu t\sqrt{2}}e^{-{1\over 2}} $$
(and the minimum one is the same with a minus sign, which is the other solution of the derivative).
2- the velocity profile for $t=1s$ is the following function of $x$: $$\omega({x, t=1s})={\Lambda x \over (4\nu)^{3\over 2}}e^{-{x^2\over (4\nu)}}$$
the velocity profile for $t=2s$ is
$$\omega({x, t=2s})={\Lambda x \over (8\nu)^{3\over 2}}e^{-{x^2\over 8\nu)}}$$
and so on.. You just have the value of $t$ inside your $\omega(x, t)$ formula. At each time there is a different profile.
Here is a sketch of different profiles at different times (I set $\Lambda=1$ and $4\nu =1 $). The bigger the lobe, the smaller the times. Selected times are $t=0.5, 1, 2$. Y axis is $\omega$: