A projectile is launched with an initial speed u such that its...

The horizontal range $R$ and maximum height $H$ of a projectile are given by:
$R = \dfrac{u^2 \sin 2\theta}{g} = \dfrac{2u^2 \sin \theta \cos \theta}{g}$
$H = \dfrac{u^2 \sin^2 \theta}{2g}$

Given that $R = H$:
$\dfrac{2u^2 \sin \theta \cos \theta}{g} = \dfrac{u^2 \sin^2 \theta}{2g}$

Simplifying this gives:
$2 \cos \theta = \dfrac{\sin \theta}{2}$
$\tan \theta = 4$

From $\tan \theta = 4$, we can find $\cos \theta$ using a right-angled triangle with opposite side $4$ and adjacent side $1$. The hypotenuse is $\sqrt{4^2 + 1^2} = \sqrt{17}$.
Thus, $\cos \theta = \dfrac{1}{\sqrt{17}}$.

The speed at the highest point is the horizontal component of the initial velocity, as the vertical component becomes zero:
$v = u \cos \theta = \dfrac{u}{\sqrt{17}}$

Answer: $\dfrac{u}{\sqrt{17}}$