That would be difficult, but in practice only due to the ramp angle - the power is there!

Neglecting air resistance, the maximum ballistic trajectory comes at a launch angle of 45 degrees on a notional flat Earth.

Time to the apex of the parabolic path is then t=V/[g root(2) ]. Total flight time is twice this: t= [V root(2)]/g

In this time, the distance travelled is L=V*root(2)*t

hence L=V^2 * 2/g

and V = root[gL/2]

A quick bash of Excel yields the following table for V vs L:

miles mph
0.0 0.0
0.0 19.9
0.0 35.3
0.1 62.9
0.3 111.8
1.0 198.8
3.2 353.5
10.0 628.6
31.6 1117.7
100.0 1987.7

So a nominal 200 mph launch on a 45 degree ramp would yield a range of ~1 mile, while "several" miles requires 350 mph.

I guess if the catapult is not speed-limited and is capable of accelerating 60,000 lb (25 tons) to 150 mph, you could get double that speed at 1/4 the load (a 6-ton pickup?)

In practice, aerodynamic drag would severely restrict the range of a non-stabilised auto body.

I'd certainly like to see the navy entry for the annual pumpkin-throwing competition.