The ball's horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] at time [tex]t[/tex] are given by
[tex]x=v_0\cos\theta\,t[/tex]
[tex]y=v_0\sin\theta\,t-\dfrac g2t^2[/tex]
where [tex]v_0=45\,\dfrac{\rm m}{\rm s}[/tex], [tex]\theta=30^\circ[/tex], and [tex]g=9.80\,\dfrac{\rm m}{\mathrm s^2}[/tex]. The ball reaches the ground when [tex]y=0[/tex] at
[tex]0=v_0\sin\theta\,t-\dfrac g2t^2=\left(v_0\sin\theta-\dfrac g2t\right)t=0\implies t=\dfrac{2v_0\sin\theta}g[/tex]
(we don't care about [tex]t=0[/tex])
At this time, the ball's horizontal position is
[tex]v_0\cos\theta\left(\dfrac{2v_0\sin\theta}g\right)=\dfrac{{v_0}^2\sin2\theta}g[/tex]
which you might recognize as the range formula. With the known parameters, the ball thus traverses a range of
[tex]\dfrac{\left(45\,\frac{\rm m}{\rm s}\right)^2\sin60^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\approx180\,\rm m[/tex]