Gun Shot Projectile

Solve $100 t \sin (\theta )-\frac{10 t^2}{2}=0$ for $t$ for time of flight to edge of building, giving $t=20 \sin (\theta)$ .

Solve $20 \sin (\theta) 100 \cos (\theta)=960$ for steepest $\theta$ for closest approach to base of building, giving $\theta =2 \tan ^{-1}\left(\frac{1}{2}\right)$ .

Solve $\frac{g t^2}{2}+960+t 100 \sin (2 \tan ^{-1}\left(\frac{1}{2}\right) )=0$ for $t$ for time of flight to base of building, giving $t=24$ .

Evaluate $100 \cos(2 \tan ^{-1}\left(\frac{1}{2}\right)$ 24) giving $1440$ . Subtracting 960 from 1440 gives 480.

The optimum angle will be as steep as possible and have the bullet just graze the edge of the building.

I'll set the person at $(0,960)$ and the building edge at $(960,960)$ . When the bullet hits the ground, subtract 960 to find the distance from the building.

Projectile equation $y=x \cdot tan{\theta}-\frac{g\cdot x^{2}}{2v_{0}\cos^{2}{\theta}}+h_{0}$ and substituting everything in allows us to solve for $\theta$

$960=960\tan{\theta}-\frac{0.48}{\cos^{2}{\theta}}+960$

$0.48 = \sin{\theta}\cos{\theta}$

$0.96= 2\sin{\theta}\cos{\theta}=\sin{2\theta}$ (note this equation has two solutions $0^{\circ}<2 \theta <180^{\circ}$ )

$2\theta=180^{\circ} - \sin^{-1}{0.96}$ (because we want the larger of the two possible angles).

Since we're about to need them, a little elementary trig gives: $\tan{\theta}=\frac{4}{3}$ and $\cos{\theta}=\frac{3}{5}$ so the projectile equation becomes

$y=-\frac{1}{720}x^{2}+\frac{4}{3}x+960$

Solving for $y=0$ is a quadratic, whose positive solution is $x=1440$ so the solution is $1440-960=\boxed{480}$