 Solution

Outside a spherical Earth the gravitational field is the field of its total mass at the center, by the shell theorem. Since $G{M}_E=g{R}^2$, the inverse-square force on the upward radial trajectory givesThe independent dimensional inputs are $R,V,g$, of dimensions $L,L{T}^{-1},L{T}^{-2}$. A dimensionless monomial in them must be a power of $V^2/(gR)$. Thus dimensional analysis givesFor a finite turning height and time, assume $0<V<\sqrt{2gR}$, equivalently $0<\lambda <2$. This escape velocity restriction is implicit in the requested maximum-height formulas.

The conservation of energy, per unit particle mass, issoThe numerator vanishes at the turning point, givingOn ascent $\dot{z}>0$, and integrating $dt=dz/\dot{z}$ givesThe upper endpoint has an integrable inverse-square-root singularity. Set $z=Hx$, use $H=R\lambda /(2-\lambda )$ and $V^{2}R=(2gR-V^2)H$, and note $R\lambda /V=V/g$. The result isFor small $\lambda$, $F(\lambda )\sim \lambda /2$ and $G(\lambda )\to 1$, recovering $H\sim V^2/(2g)$ and $T\sim V/g$. At $\lambda =2$ the particle escapes marginally and has no finite maximum; for $\lambda >2$ it escapes with nonzero asymptotic speed. The finite-turning formulas must not be extended to those cases.