$0\le d<1$
:::info
DRAG BEFORE ACCELERATION
:::
$v(t)=v_0(1-d)^t-a{1-(1-d)^t\over d}=\left(v_0d+a\over d\right)(1-d)^t-{a\over d}$
$p(t)=-at+\displaystyle\sum_{k=0}^{t-1}v(k)=\left(v_0d+a\over d^2\right)-\left(v_0d+a\over d^2\right)(1-d)^t-a\left({1+d\over d}\right)t$
$v(t_0)\le a\implies t_0\ge\log_{(1-d)}\left(a+ad\over v_0d+a\right)$
$t_0=\log_{(1-d)}\left(a+ad\over v_0d+a\right)+E\qquad0\le E<1$
$h_{max}=p(t_0)=\left(v_0d+a\over d^2\right)-a{1+d\over d}\log_{(1-d)}\left(a+ad\over v_0d+a\right)-a{1+d\over d^2}(1-d)^E-a{1+d\over d}E=$
$=a{1+d\over d}\cdot\left({v_0\over a(1+d)}+{1\over d(1+d)}-\log_{(1-d)}\left(a+ad\over v_0d+a\right)-{(1-d)^E\over d}-E\right)$
$(1-d)^\left({\left({d\over a}h_{max}-{1\over d}\right)\over(1+d)}+\left({(1-d)^E\over d}+E\right)\right)=(1-d)^{v_0\over a(1+d)}\cdot\left(v_0d+a\over a(1+d)\right)$
${\ln(1-d)\over d}(1-d)^\left({d\over a(1+d)}h_{max}+\left({(1-d)^E\over d}+E\right)\right)=\ln(1-d)\cdot(1-d)^\left({v_0\over a(1+d)}+{1\over d(1+d)}\right)\cdot\left({v_0\over a(1+d)}+{1\over d(1+d)}\right)$
$v_0=a\cdot\left({1+d\over\ln(1-d)}W\left({\ln(1-d)\over d}(1-d)^\left({d\over a(1+d)}h_{max}+\left({(1-d)^E\over d}+E\right)\right)\right)-{1\over d}\right)$
$v_0=a\cdot\left({1+d\over\ln(1-d)}W\left({\ln(1-d)\over d}(1-d)^\left({d\over a(1+d)}h_{max}+f(E)\right)\right)-{1\over d}\right)$

Additional relations:
$v_0(v,t)=\left(v+{a\over d}\right)\cdot(1-d)^{-t}-{a\over d}$
$v_0(p,t)=\left(p+a\left({1+d\over d}\right)t\right)\cdot{d\over1-(1-d)^t}-{a\over d}$
$t(v_0,v)=\log_{(1-d)}\left(\left(v+{a\over d}\right)\cdot\left(d\over v_0d+a\right)\right)=\log_{(1-d)}\left(vd+a\over v_0d+a\right)$

${v_0d+a\over ad(1+d)}=\left({v_0d+a\over ad(1+d)}-{p\cdot d\over a(1+d)}-t\right)(1-d)^{-t}$
$\ln(1-d)\cdot\left(v_0d+a\over ad(1+d)\right)(1-d)^\left({v_0d+a\over ad(1+d)}-{p\cdot d\over a(1+d)}\right)=\ln(1-d)\cdot\left({v_0d+a\over ad(1+d)}-{p\cdot d\over a(1+d)}-t\right)(1-d)^\left({v_0d+a\over ad(1+d)}-{p\cdot d\over a(1+d)}-t\right)$
$t(v_0,p)={v_0d+a\over ad(1+d)}-{p\cdot d\over a(1+d)}-{1\over\ln(1-d)}W\left(\ln(1-d)\cdot\left(v_0d+a\over ad(1+d)\right)\cdot(1-d)^\left({v_0d+a\over ad(1+d)}-{p\cdot d\over a(1+d)}\right)\right)$

$v+p\cdot d=v_0-a\cdot(1+d)\cdot t$

$v+{a\over d}=\left(v_0+{a\over d}\right)(1-d)^\left(v_0-v-p\cdot d\over a(1+d)\right)$
${\ln(1-d)\over a(1+d)}\left(v+{a\over d}\right)(1-d)^\left({v+p\cdot d\over a(1+d)}+{1\over d(1+d)}\right)=\ln(1-d)\cdot\left(v_0d+a\over ad(1+d)\right)(1-d)^\left(v_0d+a\over ad(1+d)\right)$
$v_0(v,p)={a(1+d)\over\ln(1-d)}W\left({\ln(1-d)\over a(1+d)}\left(v+{a\over d}\right)(1-d)^\left({v+p\cdot d\over a(1+d)}+{1\over d(1+d)}\right)\right)-{a\over d}$
:::info
DRAG AFTER ACCELERATION
:::
$v(t)=v_0(1-d)^t-a{1-(1-d)^t\over d}(1-d)=\left(v_0d+a-ad\over d\right)(1-d)^t-a\left(1-d\over d\right)$
$p(t)=-at+\displaystyle\sum_{k=0}^{t-1}v(k)=\left(v_0d+a-ad\over d^2\right)-\left(v_0d+a-ad\over d^2\right)(1-d)^t-{a\over d}t$
$v(t_0)\le a\implies t_0\ge\log_{(1-d)}\left(a\over v_0d+a-ad\right)$
$t_0=\log_{(1-d)}\left(a\over v_0d+a-ad\right)+E\qquad0\le E<1$
$h_{max}=p(t_0)=\left(v_0d+a-ad\over d^2\right)-{a\over d}\log_{(1-d)}\left(a\over v_0d+a-ad\right)-{a\over d^2}(1-d)^E-{a\over d}E=$
${a\over d}\cdot\left({v_0\over a}+{1-d\over d}-\log_{(1-d)}\left(a\over v_0d+a-ad\right)-{(1-d)^E\over d}-E\right)$
$(1-d)^\left({d\over a}h_{max}+{d-1\over d}+\left({(1-d)^E\over d}+E\right)\right)=(1-d)^{v_0\over a}\cdot\left(v_0d+a(1-d)\over a\right)$
${\ln(1-d)\over d}(1-d)^\left({d\over a}h_{max}+\left({(1-d)^E\over d}+E\right)\right)=\ln(1-d)\cdot(1-d)^\left({v_0\over a}+{1-d\over d}\right)\cdot\left({v_0\over a}+{1-d\over d}\right)$
$v_0=a\cdot\left({1\over\ln(1-d)}W\left({\ln(1-d)\over d}(1-d)^\left({d\over a}h_{max}+\left({(1-d)^E\over d}+E\right)\right)\right)-{1-d\over d}\right)$
$v_0=a\cdot\left({1\over\ln(1-d)}W\left({\ln(1-d)\over d}(1-d)^\left({d\over a}h_{max}+f(E)\right)\right)-{1-d\over d}\right)$

Additional relations:
$v_0(v,t)=\left(v+a\left(1-d\over d\right)\right)\cdot(1-d)^{-t}-a\left(1-d\over d\right)$
$v_0(p,t)=\left(p+{a\over d}t\right)\cdot{d\over1-(1-d)^t}-a\left(1-d\over d\right)$
$t(v_0,v)=\log_{(1-d)}\left(\left(v-a+{a\over d}\right)\cdot\left(d\over v_0d+a-ad\right)\right)=\log_{(1-d)}\left(vd+a-ad\over v_0d+a-ad\right)$

${v_0d+a\over ad}-1=\left({v_0d+a\over ad}-1-{d\over a}p-t\right)(1-d)^{-t}$
$\ln(1-d)\cdot\left({v_0d+a\over ad}-1\right)^\left({v_0d+a\over ad}-1-{d\over a}p\right)=\ln(1-d)\cdot\left({v_0d+a\over ad}-1-{d\over a}p-t\right)(1-d)^\left({v_0d+a\over ad}-1-{d\over a}p-t\right)$
$t(v_0,p)={v_0d+a\over ad}-1-{d\over a}p-{1\over\ln(1-d)}W\left(\ln(1-d)\cdot\left({v_0d+a\over ad}-1\right)\cdot(1-d)^\left({v_0d+a\over ad}-1-{d\over a}p\right)\right)$

$v+p\cdot d=v_0-a\cdot t$

$v+a\left(1-d\over d\right)=\left(v_0+a{1-d\over d}\right)(1-d)^\left(v_0-v-p\cdot d\over a\right)$
$\ln(1-d)\cdot\left({v\over a}+{1-d\over d}\right)(1-d)^\left({v+p\cdot d\over a}+{1-d\over d}\right)=\ln(1-d)\cdot\left({v_0\over a}+{1-d\over d}\right)(1-d)^\left({v_0\over a}+{1-d\over d}\right)$
$v_0={a\over\ln(1-d)}W\left(\ln(1-d)\cdot\left({v\over a}+{1-d\over d}\right)(1-d)^\left({v+p\cdot d\over a}+{1-d\over d}\right)\right)+a-{a\over d}$
:::info
POSITION NOT SLOWED DOWN (DRAG B.A.)
:::
$v(t)=v_0(1-d)^t-a{1-(1-d)^t\over d}=\left(v_0d+a\over d\right)(1-d)^t-{a\over d}$
$p(t)=\displaystyle\sum_{k=0}^{t-1}v(k)=\left(v_0d+a\over d^2\right)-\left(v_0d+a\over d^2\right)(1-d)^t-{a\over d}t$
$v(t_0)\le0\implies t_0\ge\log_{(1-d)}\left(a\over v_0d+a\right)$
$t_0=\log_{(1-d)}\left(a\over v_0d+a\right)+E\qquad0\le E<1$
$h_{max}=p(t_0)=\left(v_0d+a\over d^2\right)-{a\over d}\log_{(1-d)}\left(a\over v_0d+a\right)-{a\over d^2}(1-d)^E-{a\over d}E=$
$={a\over d}\cdot\left({v_0\over a}+{1\over d}-\log_{(1-d)}\left(a\over v_0d+a\right)-{(1-d)^E\over d}-E\right)$
$(1-d)^\left({d\over a}h_{max}-{1\over d}+\left({(1-d)^E\over d}+E\right)\right)=(1-d)^{v_0\over a}\cdot\left(v_0d+a\over a\right)$
${\ln(1-d)\over d}(1-d)^\left({d\over a}h_{max}+\left({(1-d)^E\over d}+E\right)\right)=\ln(1-d)\cdot(1-d)^\left({v_0\over a}+{1\over d}\right)\cdot\left({v_0\over a}+{1\over d}\right)$
$v_0=a\cdot\left({1\over\ln(1-d)}W\left({\ln(1-d)\over d}(1-d)^\left({d\over a}h_{max}+\left({(1-d)^E\over d}+E\right)\right)\right)-{1\over d}\right)$
$v_0=a\cdot\left({1\over\ln(1-d)}W\left({\ln(1-d)\over d}(1-d)^\left({d\over a}h_{max}+f(E)\right)\right)-{1\over d}\right)$

Additional relations:
$v_0(v,t)=\left(v+{a\over d}\right)\cdot(1-d)^{-t}-{a\over d}$
$v_0(p,t)=\left(p+{a\over d}t\right)\cdot{d\over1-(1-d)^t}-{a\over d}$
$t(v_0,v)=\log_{(1-d)}\left(\left(v+{a\over d}\right)\cdot\left(d\over v_0d+a\right)\right)=\log_{(1-d)}\left(vd+a\over v_0d+a\right)$

${v_0\over a}+{1\over d}=\left({v_0\over a}+{1\over d}-{d\over a}p-t\right)(1-d)^{-t}$
$\ln(1-d)\cdot\left({v_0\over a}+{1\over d}\right)(1-d)^\left({v_0\over a}+{1\over d}-{d\over a}p\right)=\ln(1-d)\cdot\left({v_0\over a}+{1\over d}-{d\over a}p-t\right)(1-d)^\left({v_0\over a}+{1\over d}-{d\over a}p-t\right)$
$t(v_0,p)={v_0\over a}+{1\over d}-{d\over a}p-{1\over\ln(1-d)}W\left(\ln(1-d)\cdot\left({v_0\over a}+{1\over d}\right)(1-d)^\left({v_0\over a}+{1\over d}-{d\over a}p\right)\right)$

$v+p\cdot d=v_0-at$

$v+{a\over d}=\left(v_0+{a\over d}\right)(1-d)^\left(v_0-v-p\cdot d\over a\right)$
$\ln(1-d)\cdot\left({v\over a}+{1\over d}\right)(1-d)^\left({v+p\cdot d\over a}+{1\over d}\right)=\ln(1-d)\cdot\left({v_0\over a}+{1\over d}\right)(1-d)^\left({v_0\over a}+{1\over d}\right)$
$v_0(v,p)={a\over\ln(1-d)}W\left(\ln(1-d)\cdot\left({v\over a}+{1\over d}\right)(1-d)^\left({v+p\cdot d\over a}+{1\over d}\right)\right)-{a\over d}$
:::info
POSITION NOT SLOWED DOWN (DRAG A.A.)
:::
Same formulas, use $(a\cdot(1-d))$ instead of $a$
:::info
ERROR FUNCTION
:::
$f(x)={(1-d)^x\over d}+x$
$f'(x)={\ln(1-d)\over d}(1-d)^x+1\qquad f'(x_0)\ge0\implies x_0\ge\log_{(1-d)}\left(-{d\over\ln(1-d)}\right)$
$f(0)={1\over d}\qquad f(1)={1\over d}\qquad f(x_0)=\log_{(1-d)}\left(-{d\over\ln(1-d)}\right)-{1\over\ln(1-d)}={\ln\left(-{d\over\ln(1-d)}\right)-1\over\ln(1-d)}$
$f(x_0)\le f(E)\le{1\over d}$
:::info
$d=0$ (POSITION SLOWED)
:::
$v(t)=v_0-at$
$p(t)=-at+\displaystyle\sum_{k=0}^{t-1}v(k)=v_0t-a{t(t+1)\over2}=v_0t-{a\over2}t-{a\over2}t^2=\left(v_0-{a\over2}\right)t-{a\over2}t^2$
$v(t_0)\le a\implies t_0\ge{v_0\over a}-1$
$t_0={v_0\over a}-1+E\qquad0\le E<1$
$h_{max}=p(t_0)=\left(v_0-{a\over2}\right)\left({v_0\over a}-1+E\right)-{a\over2}\left({v_0^2\over a^2}+1+E^2-2{v_0\over a}-2E+2{v_0E\over a}\right)=$
$={v_0^2\over a}-v_0+v_0E-{v_0\over2}+{a\over2}-{aE\over2}-{v_0^2\over2a}-{a\over2}-{aE^2\over2}+v_0+aE-v_0E=$
$={v_0^2\over2a}-{v_0\over2}+{aE\over2}-{aE^2\over2}={1\over2a}v_0^2-{1\over2}v_0+{a\over2}(E-E^2)={1\over2a}v_0^2-{1\over2}v_0+{a\over2}f(E)$
$v_0={a\over2}\cdot\left(1\pm\sqrt{{8\over a}h_{max}+1-4f(E)}\right)$

Additional relations:
$v_0(v,t)=v+at$
$v_0(p,t)={p\over t}+{a\over2}(t+1)$
$t(v_0,v)={v_0-v\over a}$
$t(v_0,p)={v_0-{a\over2}\pm\sqrt{\left(v_0-{a\over2}\right)^2-2ap}\over a}$

$p-vt={a\over2}(t^2-t)$
${a\over2}t^2+\left(v-{a\over2}\right)t-p=0$

$t(v,p)={{a\over2}-v\pm\sqrt{\left({a\over2}-v\right)^2+2ap}\over a}$
:::info
$d=0$ (POSITION NOT SLOWED)
:::
$v(t)=v_0-at$
$p(t)=\displaystyle\sum_{k=0}^{t-1}v(k)=v_0t-a{t(t-1)\over2}=v_0t+{a\over2}t-{a\over2}t^2=\left(v_0+{a\over2}\right)t-{a\over2}t^2$
$v(t_0)\le0\implies t_0\ge{v_0\over a}$
$t_0={v_0\over a}+E\qquad0\le E<1$
$h_{max}=p(t_0)=\left(v_0+{a\over2}\right)\left({v_0\over a}+E\right)-{a\over2}\left({v_0^2\over a^2}+E^2+2{v_0E\over a}\right)=$
$={v_0^2\over a}+v_0E+{v_0\over2}+{aE\over2}-{v_0^2\over2a}-{aE^2\over2}-v_0E=$
$={v_0^2\over2a}+{v_0\over2}+{aE\over2}-{aE^2\over2}={1\over2a}v_0^2+{1\over2}v_0+{a\over2}(E-E^2)={1\over2a}v_0^2+{1\over2}v_0+{a\over2}f(E)$
$v_0={a\over2}\cdot\left(-1\pm\sqrt{{8\over a}h_{max}+1-4f(E)}\right)$

Additional relations:
$v_0(v,t)=v+at$
$v_0(p,t)={p\over t}+{a\over2}(t-1)$
$t(v_0,v)={v_0-v\over a}$
$t(v_0,p)={v_0+{a\over2}\pm\sqrt{\left(v_0+{a\over2}\right)^2-2ap}\over a}$

$p-vt={a\over2}(t^2+t)$
${a\over2}t^2+\left(v+{a\over2}\right)t-p=0$

$t(v,p)={-v-{a\over2}\pm\sqrt{\left(v+{a\over2}\right)^2+2ap}\over a}$
:::info
ERROR FUNCTION $(d=0)$
:::
$f(x)=x-x^2$
$f'(x)=1-2x\qquad f'(x_0)\ge0\implies x_0\le{1\over2}$
$f(0)=0\qquad f(1)=0\qquad f\left({1\over2}\right)={1\over4}$
$0\le f(E)\le{1\over4}$