Two blocks A and B of masses 1 . 5   kg and 0 . 5   kg respectively are connected by a massless…

Common acceleration of the system is given by-

$\left(a_{\mathrm{sys}}\right)=\left(\frac{1.5-0.5}{1.5+0.5}\right) g=\frac{g}{2}=5 \mathrm{m} {\mathrm{s}}^{-2}$

When block of mass $1.5 \mathrm{kg}$ hits the ground, at that time block of mass $0.5 \mathrm{kg}$ will be at a height of $80 \mathrm{cm}$ from the ground. After this point string become slack and block $B$ will move with deceleration $g=10 \mathrm{m} {\mathrm{s}}^{-2}.$

Using third equation of motion, speed of block $B,$ When  block $A$ Reaches the ground-

$v^2=0+2 \left(\frac{g}{2}\right) \left(\frac{80}{100}\right)=8$

$\Rightarrow v=\sqrt{8} \mathrm{m} {\mathrm{s}}^{-1}$

Again using third equation of motion, suppose block $B$ will further move upto a height of $h \mathrm{m}.$

$0=v^2+2 \left(-10\right) h$

$\Rightarrow h=\frac{v^2}{20}=0.4 \mathrm{m}=40 \mathrm{cm}$

hence, total height of block $B$ from the ground

$=80+40=120 \mathrm{cm}$