The shell's horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] are given by
[tex]x=v_0\cos55.0^\circ\,t[/tex]
[tex]y=v_0\sin55.0^\circ\,t-\dfrac g2t^2[/tex]
where [tex]v_0[/tex] is the given speed of 1.70 x 10^4 m/s, and [tex]g[/tex] is the acceleration due to gravity (taken here to be 9.80 m/s^2).
To find the horizontal range, you can use the range formula,
[tex]x_{\rm max}=\dfrac{{v_0}^2\sin2\theta}g[/tex]
with [tex]\theta[/tex] being the angle at which the shell is fired.
Alternatively, we can work backwards and deal with part (b) first:
(b) The time spent in the air is the time it takes for the shell to reach the ground. To find that, you solve for [tex]t[/tex] in [tex]y=0[/tex]:
[tex]v_0\sin55.0^\circ\,t-\dfrac g2t^2=0\implies t\approx284\,\rm s[/tex]
(a) After this time, the shell will have traveled horizontally
[tex]x=v_0\cos55.0^\circ(284\,\rm s)\approx2.77\times10^5\,\rm m[/tex]
