Question

The cannon on a battleship can fire a shell a maximum distance of 26.0 km.

Answer 1

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