Answer: a) 0.78 s; b) 899.83 m; c) Δx=1150*Δt
Explanation: In order to solve this problem we have to use the kinematic equations for the independent two axis (x-y). The following expressions are:
y=yo+voy-(g*t^2)/2 g is 9.8 the aceletarion of the gravity on the y axis
vfy=voy-g*t
x=xo+vox+(a*t^2)/2 a=o there is not acceleration in the x axis.
vfx=vox+a*t
voy and vox ara the initial velocities for each axis.
yo and xo are the initial positions of teh bullet.
From these equations we can obtain the time
y=0=3-(g*t^2)/2
t^2=3*2/g
t=

t= 0.78 s
To calculate xf= voy*t= 1150 * 0.78= 899.83 m
Finally considering with the last equation we ontain the error in Δx
in the form:
Δx=1150*Δt since is linear the dependance with time.