it is same 100 m/sec

v = final velocity u = initial velocity a = acceleration

S = distance traveled t = time traveled

suppose at the top point it reaches, its height is H.

A body goes up to a height during the upward journey

v = u + at => 0 = u - g t => t = u/g : time to reach the top

v² - u² = 2 a S => 0² - u² = 2 (-g) H => u² = 2 g H => H = 1/2u²/g

OR, H = ut + 1/2 a t² = u*u/g -1/2 g (u/g)² = u²/g - 1/2 u²/g

So H = 1/2 u² / g ------ equation 1

A body falls freely from height H: then

v = u+at => v = 0+ g t = gt

v² - u² = 2 a S => v² - 0 = 2 g S => S = 1/2 v² / g

If the body travels a distance H on the way down then, above formula is:

H = 1/2 v² / g --- which is same as equation 1

So velocity at a height H from ground on the way up

= velocity at same height H from ground on the way down