A Stone dropped from top of a well reaches the surface of water in 2 seconds, find the velocity of stone while it touches the surface of water and what is the depth of the water surface from top of well (`g=10m//s^(2)`) (Using V = U + at, S = Ut + `1//2at^(2)`)

Explanation

Given that
t = 2s
u = 0 m/s [`because` free fall body]
v = ?
Depth s = ?
a = g = 10`m//s^(2)`
i) v = u + at
`v=0+10xx2=20m//s`
ii) `s=ut+1/2at^(2)`
`=0+1/2xx10xx2^(2)`
`=1/2xx10xx4`
= 20m
Hence, velocity of stone while it touches the surface of water = 20 m/s
Depth of the water surface from the top of well = 20m.