Answer
a) $y=\frac{1}{2}x$
b) Straight line.
c) $1088\;\rm m$
Work Step by Step
a)
To find $y_{(x)}$, we need to find $a_x$ and $a_y$ first.
Applying Newton's second law on the rocket in both directions:
$$\sum F_x=F_T\cos\theta=ma_x$$
where $F_T$ is the thrust force.
Thus,
$$a_x=\dfrac{F_T\cos\theta}{m}=\dfrac{140700\cos44.7^\circ}{5000}$$
$$a_x=\bf 20.0\;\rm m/s^2$$
$$\sum F_y=F_T\sin\theta-mg=ma_y$$
Thus,
$$a_y=\dfrac{F_T\sin\theta-mg}{m}=\dfrac{140700\sin44.7^\circ-5000\times 9.8}{5000}$$
$$a_y=\bf 9.99\;\rm m/s^2$$
Its horizontal displacement is given by
$$x=x_0+v_{0x}t+\frac{1}{2}a_xt^2$$
Plugging the known; where we know that the rocket starts from rest.
$$x=0+(0)t+\frac{1}{2}a_xt^2=\frac{1}{2}\times 20t^2$$
$$x=10t^2\tag 1$$
Its vertical displacement is given by
$$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$$
Plugging the known; where we know that the rocket starts from rest.
$$y=0+(0)t+\frac{1}{2}a_yt^2=\frac{1}{2}\times 9.99 t^2$$
$$y=4.995 t^2\approx 5t^2\tag 2$$
Solving (1) for $t^2$ and plugging the result into (2);
from (1) $$t^2=\dfrac{x}{10}$$
Thus,
$$y= 5t^2=5\times \dfrac{x}{10} $$
$$\boxed{y= \frac{1}{2}x }$$
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b)
The shape of the trajectory is given by the boxed equation above which is a straight line, as we see in the figure below.
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c)
We need to find the height when the rocket's speed is $v_s=$330 m/s.
Since the rocket is moving in a straight line, as we found in part [b] above, the speed direction is constant.
So that when the velocity of the rocket reaches 330 m/s, its vertical velocity component will be given by
$$v_y=\overbrace{v_{0y}}^{=0}+a_yt=9.99t$$
And the horizontal velocity component will be given by
$$v_x=\overbrace{v_{0x}}^{=0}+a_xt=20t$$
Thus,
$$v_s^2=v_x^2+v_y^2$$
$$330^2=(20t)^2+(9.99t)^2$$
Solving for $t$;
$$330^2= t^2(20^2+9.99^2)$$
$$t=\sqrt{\dfrac{330^2}{20^2+9.99^2}}=\bf 14.76\;\rm s$$
Now we can use the vertical elevation at $v=v_s$ using equation (2);
$$y=4.995 t^2= 4.995 \times14.76^2=\color{red}{\bf 1088}\;\rm m $$
