Question

Stone dropped in a well hits water surface after 2 sec. What is the depth of well and with what speed will the stone hit the water surface?

Answer 1

U = 0
t = 2sec
a = -g = 9.8m/s^2

S = ut + 1/2at^2
S = 0 + (-19.6)
S = -19.6m ( depth

v = u + at
v = -19.6 m/s

as body is falling from a height so , depth of the well will be 19.6m

and speed with which it strikes the surface of water = 19.6m/s