The projectile's horizontal and vertical positions at time [tex]t[/tex] are given by
[tex]x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t[/tex]
[tex]y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2[/tex]
where [tex]g=9.8\dfrac{\rm m}{\mathrm s^2}[/tex]. Solve [tex]y=0[/tex] for the time [tex]t[/tex] it takes for the projectile to reach the ground:
[tex]30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s[/tex]
In this time, the projectile will have traveled horizontally a distance of
[tex]x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m[/tex]
The projectile's horizontal and vertical velocities are given by
[tex]v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ[/tex]
[tex]v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt[/tex]
At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about [tex]\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}[/tex].
