It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is [tex]\theta = 45^{\circ}[/tex].
In fact, the laws of motions on both x- and y- directions are
[tex]S_x(t)= v_0 cos \theta t[/tex]
[tex]S_y(t)= v_0 \sin \theta t - \frac{1}{2} gt^2[/tex]
From the second equation, we get the time t at which the projectile hits the ground, by requiring [tex]S_y(t)=0 [/tex], and we get:
[tex]t= \frac{2 v_0 \sin \theta}{g} [/tex]
And inserting this value into Sx(t), we find
[tex]S_x(t) = 2 \frac{v_0^2}{g} \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)[/tex]
And this value is maximum when [tex]\theta=45^{\circ}[/tex], so this is the angle at which the projectile reaches its maximum distance.
So now we can take again the law of motion on the x-axis
[tex]S_x(t)= \frac{v_0^2}{g} \sin (2\theta)[/tex]
And by using [tex]S_x = 26 km=26000 m[/tex], we find the value of the initial velocity v0:
[tex]v_0 = \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s[/tex]
