This question is incomplete, the missing image is uploaded along this answer below.
Answer:
the speed of the 50-kg cylinder after it has descended is 3.67 m/s
Explanation:
  Given the data in the question and the image below;
relation between velocity of cylinder and velocity of the drum is;
V = ω
 = ω × r
 × r ----- let this be equ 1
  ----- let this be equ 1
where V is velocity of cylinder,  ω
 is velocity of cylinder,  ω is the angular velocity of drum C and r
 is the angular velocity of drum C and r is the radius of drum C
 is the radius of drum C
Now, Angular velocity of gear B is;
ω = ω
 = ω
ω = V
 = V / r
 / r -------- let this equ 2
  -------- let this equ 2
so;
V / 0.1 m = 10V
 / 0.1 m = 10V
Next, we determine the angular velocity of gear A;
from the diagram;
ω ( 0.15 m ) = ω
( 0.15 m ) = ω ( 0.2 m )
( 0.2 m )
from equation 2; ω = V
 = V / r
 / r 
 
so
ω ( 0.15 m ) = (V
( 0.15 m ) = (V / r
 / r ) 0.2 m
 ) 0.2 m 
substitutive in value of radius r (0.1 m)
 (0.1 m)
ω ( 0.15 m ) = (V
( 0.15 m ) = (V / 0.1 m ) 0.2 m
 / 0.1 m ) 0.2 m 
ω ( 0.15 ) = 0.2V
( 0.15 ) = 0.2V / 0.1
 / 0.1 
ω =  2V
 =  2V / 0.15
  / 0.15
ω = 13.333V
 = 13.333V ----- let this be equation 3
   ----- let this be equation 3
To get the speed of the cylinder, we use energy conversation;
assuming that the final position is;
T₁ + ∑ = T₂
 = T₂
0 + m gh =
gh =  m
m V²
V² +
 +  ω²
ω² +
 +  ω²
ω² 
 
so
m gh =
gh =  m
m V²
V² +
 +  (m
(m k
k ²)(13.333V
²)(13.333V )² +
)² +  (m
(m k
k ²)(10V
²)(10V )²
)²
we given that; m = 50 kg, h = 2 m, m
 = 50 kg, h = 2 m, m = 10 kg, k
 = 10 kg, k 125 mm = 0.125 m, m
 125 mm = 0.125 m, m = 30 kg, k
 = 30 kg, k = 150 mm = 0.15 m.
 = 150 mm = 0.15 m.
we know that; g = 9.81 m/s²
so we substitute
50 × 9.81 × 2 = (  × 50 × V
 × 50 × V ²) +
²) +  ( 10 × (0.125)² )(13.333V
( 10 × (0.125)² )(13.333V )² +
)² +  ( 30 × (0.15)²)(10V
( 30 × (0.15)²)(10V )²
)²
981 = 25V ² + 13.888V
² + 13.888V ² + 33.75V
² + 33.75V ²
²
981 = 72.638V ²
²
V ² = 981 / 72.638
² = 981 / 72.638
V ² = 13.5053
² = 13.5053
V = √13.5053
 = √13.5053
V = 3.674955 ≈ 3.67 m/s
 = 3.674955 ≈ 3.67 m/s
Therefore,  the speed of the 50-kg cylinder after it has descended is 3.67 m/s