Answer
a) $165\;\rm m$
b) Straight line.
Work Step by Step
a) Since the thrust is constant and is horizontal through the whole trip and since air resistance is negligible here, the only horizontal force exerted on the rocket is the thrust force.
Thus,
$$\sum F_x=F_T=ma_x$$
where $F_T$ is the thrust force.
Thus,
$$a_x=\dfrac{F_T}{m}$$
Plugging the known;
$$a_x=\dfrac{20}{500\times 10^{-3}}=\bf 40\;\rm m/s^2\tag 1$$
Now we need to find the time it takes to touch the ground so we can find the horizontal displacement using this time.
$$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$$
The initial height is 40 m and the initial vertical velocity component is zero since the rocket is lunched horizontally. Also, its vertical acceleration component is equal to the free-fall acceleration since there is no vertical forces exerted on it except its own weight due to Earth's gravitational pull.
$$0=40+(0)t-\frac{1}{2}gt^2$$
Thus,
$$t =\sqrt{\dfrac{2\times 40}{9.8}}=\bf 2.86\;\rm s$$
So,
$$x=x_0+v_{0x}t+\frac{1}{2}a_xt^2$$
Plugging the known;
$$x=0+(0.5)(2.86)+\frac{1}{2}(40)(2.86)^2$$
$$x=\color{red}{\bf 165}\;\rm m$$
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b) To describe the trajectory of the rocket we need to find $y_{(x)}$.
We know that
$$y=40-4.9t^2\tag 1 $$
while
$$x=0.5t+20t^2\approx 20t^2\tag 2$$
And since for an interval time $20t^2$ is much bigger than $0.5t$, we can approximate this equation as we did above.
Plugging $t^2$ from (2) into (1);
$$\boxed{y=40-\dfrac{4.9x}{20}} $$
And then the trajectory is, as seen below, a straight line.
