It is given that, mA=1.5kg,mB=0.5kg The common acceleration of two masses is a‌‌=(‌
mA−mB
mA+mB
)g ‌‌=(‌
1.5−0.5
1.5+0.5
)×10 ‌‌=5m∕s2 The figure below represents the block A is lifted until block touches the ground.
From the equation of motion, v2=u2+2gh v2=0+2(10)h v2=20h For initial condition, the block A falls at the height of 0.8m vi2=ui2−2a(0.8) 0=v2−2(5)(0.8) Substitute the values in the above equation. v2=2(5)(0.8) 20h=2(5)(0.8) h=0.4m The maximum height reached by the block B is hmax=0.8+0.4=1.2m hmax=120cm