Velocity of the body on reaching the ground is same in magnitude in the following cases
a) a body projected vertically from the top of tower of height 'h' with velocity 'u'
b) a body thrown down wards with velocity 'u' from the top of tower of height 'h'
c) a body projected horizontally with a velocity 'u' from the top of tower height 'h'
d) a body dropped from the top tower of height 'h'

For the body projected vertically upwards we can write ,

V² = (-u)² + 2gh = u² +2gh

For the body projected vertically downwards we can write ,

V² = u² +2gh

For the body projected horizontally with velocity 'u' , horizontal component of velocity just before it strikes the groud is V_h = u

And vertical component of velocity just before it strikes the ground is

V_v² = 0 + 2gh

.: If 'v' be the speed of the projectile just before it strikes the ground is given by ,

V² = V_h² +V_v² = u² + 2gh

For the body dropped from the top of a tower of height h , its speed v just before it strikes the ground is given by

V² = 0 +2gh

So ,option (2) is correct.