You shoot a model rocket into the air with a speed of 18.2 m/s. now high does the rocket go?

From 3rd equation of motion we get

v² = u² + 2gh

as final velocity is 0 m/s; accleration due to gravity = 9.8 m/s²

(0) ² = (18.2) ²+2 (-9.8) (h)

0=331.24 - 19.6 (h)

(-331.24) : (-19.6) = h

therefore h = 16.9 meters

We don't know anything about the engine on the model rocket ...

how long it burns after launch, how much force it exerts, how much

impulse or acceleration it imparts to the rocket, etc.

The only thing we can do with this question is to think of the craft as

a rock instead of a rocket ... assume that after launch, there are no

forces on it except the force of gravity.

Acceleration of gravity = 9.8 m/s²

This is the amount of speed lost each second.

So the rocket continues to climb for (18.2) / (9.8) = 1.857 seconds

At the bottom, its speed is 18.2.

At the top, its speed is zero.

Its average speed during the climb is (1/2) (18.2 + 0) = 9.2 m/s.

It climbs at an average speed of 9.2 m/s for 1.857 seconds.

It reaches the altitude of (9.2) x (1.857) = 17.09 meters above the launch pad.