Assume that the magnitude and direction due to gravity at a point outside Earth at distance $x$ from the Earth's centre is equal to $-\frac{k}{x^2}$, where $k$ is a constant.
i) Neglecting atmospheric resistance, prove that is an object is projected from the Earth's surface with speed $u$, it's speed $v$ in any position is given by $v^2=u^2-2gR^2(\frac1R-\frac1x)$ where $R$ is the Earth's radius and $g$ is the magnitude of acceleration due to Earth's gravity at the Earth's surface.
ii) Show that the greatest height $H$ above the Earth's surface is given by $H=\frac{u^2R}{2gR-u^2}$
iii) Hence if the radius of the Earth is $6400$km and the acceleration due to gravity is $9.8ms^{-1}$ find the speed required by the object to escape Earth's gravitational influence.
I've done part i) and iii), I'm struggling with part ii)
For Part ii) I've let $v=0$ but I get $x=\frac{2gR^2}{2gR-u^2}$
I'm pretty sure you made this mistake: Substituting $H=x$ instead of $R+H=x$. With $x=R+H$, and with $v=0$, we have $u^2=2gR^2\left(\dfrac{1}{R}-\dfrac{1}{R+H}\right)=\dfrac{2gRH}{R+H} \implies H=\dfrac{u^2R}{2gR-u^2}$. Hope it helps!