Answer
The bullet leaves the gun at an angle:
$\theta =\tan ^{-1}{\ }(\frac{b}{a})$.
The bullet lands at a distance $x=(a b)/16$ from the origin.
Work Step by Step
The slope of the tangent line is $\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{b-32 t}{a}$.
The bullet leaves the gun at $t=0$, so the slope of the tangent line at $t=0$ is $\frac{dy}{dx}|_{t=0}=\frac{b}{a}$.
Since the slope of tangent line is equal to $\tan \theta$, where $\theta$ is the angle between the line and the $x$-axis, we have $\tan \theta=b/a$. So, $\theta =\tan ^{-1}{\ }(\frac{b}{a})$.
The bullet lands at a distance when $y=0$. So we solve the equation $y=b t-16t^2=0$.
$t(b-16t)=0$,
The solutions are $t=0$ or $t=b/16$. Since $t=0$ corresponds to the time when the bullet leaves the gun, $t=b/16$ is the time when it lands on the ground. Substituting $t=b/16$ into $x$ we get the distance from the origin: $x(b/16)=(a b)/16$.
