A stone that starts at rest is in free fall for 4.0 seconds. What's the stone’s velocity after 4.0 seconds?

 and stone’s displacement during this time? 
Physics

oldschool2021-04-07T16:14:56Z

Vf = Vi + a*t = 0 + 9.8*4 = 39m/s
d = do + Vo*t +½*a*t² = 0 + 0*t +½*9.8*4² = 78.4 or 78m with 2 s.f. 

Morningfox2021-04-07T15:55:20Z

Each second, the stone gains speed of 9.8 m/s, downwards.

oubaas2021-04-07T14:54:11Z

final velocity V = g*t = -9.806*4 = -39.2 m/sec 

displacement d = V*t/2 = g*4*4/2 = g*8 = -9.806*8 = -78.4 m/sec 
displacement d = g/2*t^2 = g*16/2 = g*8 = -9,806*8 = -78.4 m 
displacement d = V^2/2g = g^2*16/2g = g*8 = -9.806*8 = -78.4 m