I asked such a question before but I do learn best by mistakes and corrections.(I didn't fully understand it yet.) I could really use your verification: A cannonball is being fired with a velocity of 800 $m\over s$ in a straight line with 60 degrees from the ground. What is its range?
Attempt: $v_0=(800\cos 60,800\sin60)$. $a=a_0=-g=(0,-g)$. $v(t)=800\cos60, 800\sin60-gt)$ and $x(t)=(800\cos60t,800\sin60t-{g\over2}t^2)$. The range is the horizontal distance the cannonball travels.(Is it? I am not sure.) In the point it stops, the y-component of the body is 0. By simple computing, $t={400\sqrt{3}\over g}$. The range(?) is, if so, $800\cos60\cdot t=400\cdot {400\sqrt{3}\over g}$.
I'm gonna put in a long answer here since I forgot all of my kinematic equations (sad I know).
From what I remember from taking a course on fundamental physics in college, acceleration is just the second derivative of displacement with respect to time or the first derivative of velocity $\vec a=\frac{d^2 \vec s}{dt^2} = \frac {dv}{dt}$ where $s$ is the displacement.
Thus doing the integrals:
$$ \int \vec a \, dt = \int d \vec v$$ $$ \vec at + \vec v_0 = \vec v(t) \tag{1}\label{1} $$ $$ \int (\vec a t + \vec v_0)\, dt = \int d \vec s$$ $$ \frac {1}{2}\vec at^2 + \vec v_0t + \vec s_0 = \vec s(t) \tag{2}\label{2}$$
Obtaining the needed kinematic equations, we will then solve the problem. Knowing that the cannonball will stop when it lands, we will just have to find the time $t$ when the cannonball is in the ground again, thus we would have to set the displacement in the y-direction $s_y(t)$ to be zero and find the value of time from there. Then using time $t$ and velocity $v_x$, we will plug it in back to $\eqref{2}$ to get the displacement.
$$ 0 = \vec s(t) = v_0yt + \frac{1}{2}\vec a t^2 = t(v_y + \frac{1}{2} \vec a t)$$
We will get two values for $t$ here, one is at the very instant the the cannon fired $t = 0$ and the other one is when the cannonball landed at the ground. Of course we would look for the non-trivial solution so that we would have a good time of solving this.
$$ -v_y = \frac{1}{2} \vec a t $$ $$ \frac {-2 v_y}{\vec a} = t \tag{3}\label{3}$$
We now have the value of $t$ when $s_y(t) = 0$ so we will just have to plug it into equation \eqref{2} for $s_x(t)$. For simplicity's sake, let's call the obtained value of $t$ as $t_1$.
$$\eqref{3} \, \text{to} \, \eqref{2}$$ $$s_x(t_1) = v_0x(t)$$ $$s_x(t_1) = v_0x\left(\frac{-2v_y}{\vec a}\right)$$
So for completeness sake, let's compute for the numerical answer. Letting $a=g=-9.81 \frac {m}{s^2}$, $| \vec v_0| = 800 \frac ms$ and $\vec v_0 = \langle 800\cos60, 800\sin60 \rangle$, we would then compute:
$$ s_x(t_1) = 800 \cos 60 \, \frac {m}{s}\left(\frac {-2}{-9.81 \frac{m}{s^2}}\right)800 \sin60 \, \frac{m}{s}$$ $$ s_x(t_1) = \frac {800}{2} \, \frac{m}{s} \left(\frac {2}{9.81 \frac{m}{s^2}} \right) \frac {800}{2}\sqrt{3} \, \frac{m}{s}$$ $$ s_x(t_1) = 400 \left(\frac {800 \sqrt{3}}{9.81} \right) \, \frac{m}{s} $$ $$ s_x(t_1) = 400 \left(\frac {800 \sqrt{3}}{g} \right) \, \frac{m}{s} $$
So you're just missing a factor of two from your answer, you might have made that mistake in obtaining the value of $t_1$ as shown earlier.