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