The answer would be a positive charge
distance to the star Betelgeuse: 640 ly
As we know that

also we know that


So the distance of Betelgeuse = 640 ly

distance to the star VY Canis Majoris: 


distance to the galaxy Large Magellanic Cloud: 49976 pc


now we have


distance to Neptune at the farthest: 4.7 billion km

now the order of distance from least to greatest is as following
1. distance to Neptune at the farthest
2. distance of Betelgeuse
3. distance to the star VY Canis Majoris
4. distance to the galaxy Large Magellanic Cloud
The planet is represented as Saturn ♄
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m