Block's colliding at some height!

let $M$ be the mass at the top and $m=\frac{M}{2}$

using conservation of momentum

$M\sqrt{2gh}=MV+mv$

$g=10~m/s^2$ , $h=0.2~m$ and $V$ and $v$ are velocities after collision.

$2=V+\frac{v}{2}$

using coefficient of restitution

$-2=V-v$

Solving

we get

$v=\frac{8}{3}~m/s$ and $V=\frac{2}{3}~m/s$

Since the vertical component in both the cases of blocks falling is zero so they reach the ground in same time.

$h^{\prime}=0+\frac{1}{2}gt^2$

Distance covered by small cube $y=vt$

Distance covered by big cube $x=Vt$

$\huge{\frac{y}{x}=4}$

Note that $V$ and $v$ are along horizontal direction.

$V~velocvity~before~strike=\sqrt{2*g*h}=\sqrt{2*9.81*0.2}=1.98.\\After~strike,~m~has~velocity~v~and ~small, ~u.\\Equating ~momentums~~before~and~after,~mV=mv+\dfrac 1 2 *m*u\\\implies~\dfrac 1 2 *u=1.98-v.......(1) \\ Equating ~Energy~before~and~after,\\m*g*h=K.E_m+K.E_{m/2}~~\implies~m*9.8*.2\\=\dfrac 1 2 *m*v^2+\dfrac 1 2 *\dfrac m 2 *u^2\\From~ (1)~1.96 = \dfrac 1 2 *v^2 +(1.98-v)^2. ~~ Solving ~for~v,~v=.99~~or~~.66. \\If~v=.99, from (1)~u=1.98~not~posible.~ So~ v=.66~from~(1), u=2.64.\\Since~both~fall~through~same~hihgt~both~are~in~air~for~same~time.\\So~horizontal~distance~covered~will~be~prpotional ~to~velocities.\\\therefore~\dfrac y x =\dfrac {2.64} { .66} =~~~~\Large \color{#D61F06}{4}$