Let the constant horizontal acceleration be #-a# (the negative sign signifying that it is opposing the motion#. The vertical and horizontal components of motion gives
#y = v_0sin theta t-1/2 g t^2 #
#x = v_0cos theta t-1/2 a t^2 #
The time of flight can be found by solving for #y=0#,
#v_0sin theta t-1/2 g t^2 =0 implies t = {2v_0 sin theta}/g#
(the other root, #t=0#, obviously denotes the time at which the projectile is launched). The value of the horizontal range is the value of #x# at this time :
#R = v_0 cos theta times {2v_0 sin theta}/g -1/2 a({2v_0 sin theta}/g)^2#
# qquad = v_0^2/{g} (2sin theta cos theta - a/g 2sin^2 theta)#
# qquad=v_0^2/{g} (sin(2theta)+a/gcos(2theta)-a/g)#
#qquad = v_0^2/{g^2} sqrt{a^2+g^2}(g/sqrt{a^2+g^2}sin(2theta)+a/sqrt{a^2+g^2}cos(2theta))-v_0^2a/g^2 #
#qquad = v_0^2/{g^2} sqrt{a^2+g^2}sin(2theta+phi)-v_0^2a/{g^2} , qquad qquad phi = tan^-1(a/g)#
Thus the range will attain its maximum value at
#2theta +phi = pi/2 implies theta = pi/4-1/2 tan^-1(a/g)#
and the maximum range is #v_0^2/{g^2} (sqrt{a^2+g^2}-a)#
