Starting from the assumption that for the time traveled, the bullet has no loss in its speed and considering the cinematic equations of movement description we have to
[tex]v = \frac{x}{t}[/tex]
Where,
x = Displacement
t = Time
Replacing with our values and rearranging to find the displacement we have
[tex]x = v*t\\x = 11*2.15\\x = 161.7m[/tex]
The distance from the base of the bridge where it lands is 161.7m
