Question

a stone dropped in a well hits the water surface after 2s . what is the depth of the the well and with what speed will the stone hit the water surface ? assume the uniform acceleration due to gravity to be 9.8m/s^2.

Answer 1

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

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

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

Answer 2

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