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`        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??`
7 years ago

510 Points
```										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
```
7 years ago
Fawz Naim
37 Points
```										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
```
7 years ago
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