Answer:
a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d)  the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
  
Explanation: 
Given that;
 height of the ramp h1 = 0.40 m
foot of the ramp above the floor h2 = 1.50 m
assuming R = 15 mm = 0.015 m
density of steel = 7.8 g/cm³
density of aluminum =  2.7 g/cm³
a) distance that the solid steel sphere sliding down the ramp without friction;
we know that
distance = speed × time
d = vt --------let this be equ 1
according to the law of conservation of energy
mgh₁ =  mv²
 mv²
v² = 2gh₁   
v = √(2gh₁)
from the second equation; s = ut +   at²
 at²
that is; t = √(2h₂/g)
so we substitute for equations into equation 1
d = √(2gh₁) × √(2h₂/g)
d = √(2gh₁) × √(2h₂/g)
d = 2√( h₁h₂ )     
we plug in our values 
d = 2√( 0.40 × 1.5 )
d = 1.55 m
Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b)
distance that a solid steel sphere rolling down the ramp without slipping;
we know that;
mgh₁ =  mv² +
 mv² +  
  ω²
ω²
mgh₁ =  mv² +
 mv² +  (
 ( mR²) ω²
mR²) ω²
v = √(  gh₁  )
gh₁  )
so we substitute √(  gh₁  ) for v and  t = √(2h₂/g) in equation 1;
gh₁  ) for v and  t = √(2h₂/g) in equation 1;
d = vt
d = √(  gh₁  ) × √(2h₂/g)
gh₁  ) × √(2h₂/g)  
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )   
d = 1.31 m
Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
  
c) 
distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;
we know that;
mgh₁ =  mv² +
 mv² +  
  ω²
ω²
mgh₁ =  mv² +
 mv² +  (
 ( mR²) ω²
mR²) ω²
v = √(  gh₁ )
gh₁ )
so we substitute √(  gh₁ ) for v and t = √(2h₂/g) in equation 1 again
gh₁ ) for v and t = √(2h₂/g) in equation 1 again
d = vt
d = √(  gh₁ ) × √(2h₂/g)
gh₁ ) × √(2h₂/g)
d = 1.549√( h₁h₂ )
d = 1.549√( 0.4 × 1.5 )
d = 1.2 m
Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) distance that a solid aluminum sphere rolling down the ramp without slipping.
we know that;
mgh₁ =  mv² +
 mv² +  
  ω²
ω²
mgh₁ =  mv² +
 mv² +  (
 ( mR²) ω²
mR²) ω²
v = √(  gh₁  )
gh₁  )
so we substitute √(  gh₁  ) for v and  t = √(2h₂/g) in equation 1;
gh₁  ) for v and  t = √(2h₂/g) in equation 1;
d = vt
d = √(  gh₁  ) × √(2h₂/g)
gh₁  ) × √(2h₂/g)  
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )   
d = 1.31 m
Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m