a man of 85 kg lowers himself from 10 m height holding a rope running over a pulley to a 65kg sand bag find his speed which hits the ground??

accleration of system = a

a = g(m₂-m₁)/(m₁+m₂)

 a  =g(85-65)/(85+65)=4/3 m/s²

now using ,

                v² = u² +2as

initial velocity = 0

      v² = 2*(4/3)*10                                     (s = 10m)

      v=5.16m/s

Let the tension in the rope be T

using Newton's second law of motion

85g-T=85a ....1

T-65g=65a ....2

solve the two equations to get the value of 'a' 

now distance=10m

using third equation of motion

v^2=u^2+2aS

u=0

v^2=2aS

put the value of a and S to get velocity