I asked a question on pen and paper apogee estimation and got an answer from @tom-spilker that called my attention: He said that he quickly estimated an inclined launch apogee from the results of a 1-D simulation. So I tried to find a general expression that could illustrate what I think that he did and I would like some feedback on whether or not this makes any sense. Here is what I did:

A one dimensional scenario is a particular case of a two dimensional scenario so we can define the acceleration vector as $\overrightarrow{a} =\begin{bmatrix} \frac{D_{x}}{M} \cdot \widehat{i}\\ (-g+\frac{D_{y}}{M}) \cdot \widehat{j} \end{bmatrix}$, where $D_{x} = D\cdot \cos \theta$ and $D_{y} = D\cdot \sin \theta$. Then, to estimate the apogee we integrate the acceleration over time to find the velocity vector, solve for the time $t$ when the vertical component is zero and after integrating the velocity vector a second time to obtain the spatial coordinates vector, we use the $t$ we have found to calculate the apogee. Now, let us go over this process symbolically only for the vertical component of the acceleration:

$$\frac{\mathrm{d} v_{y}}{\mathrm{d} t} = -g+\frac{D_{y}}{M}$$

$$\int_{t_{0}}^{t}dv_{y} = -\int_{t_{0}}^{t}g\cdot dt+\int_{t_{0}}^{t}\frac{D\cdot \sin \theta}{M}\cdot dt\;$$

$$\frac{\mathrm{d} h}{\mathrm{d} t} = -g\cdot t+\sin \theta\cdot \int_{t_{0}}^{t}\frac{D}{M}\cdot dt\; +v_{y0}$$

$$\int_{t_{0}}^{t}dh = -g\cdot \int_{t_{0}}^{t}t\cdot dt+\sin \theta\cdot \iint_{t_{0}}^{t}\frac{D}{M}dt\;+\int_{t_{0}}^{t}v_{y0}\cdot dt$$

And we finally arrive at the expression for the rocket altitude $h$ at a time instant $t$. Note first that for $v_{y}(t_{0})$ I set a initial vertical speed $v_{y0}$ (same case for $h_{0}$), and secondly that I assumed the attack angle $\theta$ remains constant. (obs: I know this is not accurate but this was the path I found to keep it "pen and paper" friendly)

$$h = -g\cdot \frac{t^{2}}{2}+\sin \theta\cdot \iint_{t_{0}}^{t}\frac{D}{M}dt\;+v_{y0}\cdot t +h_{0}$$

So if we consider the vertical launch case $h_{90^{\circ}}$ and the inclined launch case $h_{75^{\circ}}$, we can estimate that $$h_{75^{\circ}} = (h_{90^{\circ}} -(-g\cdot \frac{t^{2}}{2}+v_{y0}\cdot t +h_{0}))\cdot \sin 75^{\circ}+(-g\cdot \frac{t^{2}}{2}+v_{y0}\cdot t +h_{0}).$$

Or alternatively:

$$h_{75^{\circ}} = h_{90^{\circ}}\cdot \sin 75^{\circ} +(1 -\sin 75^{\circ})(-g\cdot \frac{t^{2}}{2}+v_{y0}\cdot t +h_{0})$$

Now, we need to find a relationship between $t_{75^{\circ}}$ and $t_{90^{\circ}}$ the apogee instants for the vertical and inclined flights.

For initial vertical speed equal to zero we have: $v_{y90^{\circ}} = -g\cdot t +\sin 90^{\circ}\cdot \int_{t_{0}}^{t}\frac{D}{M}\cdot dt$. If we write the same expression for $v_{y75^{\circ}}$ and solve for $t$ to find the apogee instant, it will become evident that $t_{apogee 75^{\circ}} = \sin 75^{\circ}\cdot t_{apogee 90^{\circ}}$.

So rewriting the apogee extimation expression, we can estimate that: $$h_{75^{\circ}} = h_{90^{\circ}}\cdot \sin 75^{\circ} +(1 -\sin 75^{\circ})(-g\cdot \frac{\sin^{2} 75^{\circ}\cdot t_{apogee\, 90^{\circ}}^{2}}{2}) $$(considerng the initial conditions to be zero.)


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