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v_{0}=0 m/s

t=3.41s

acceleration rate is a=g=9.8 m/s^{2}

v_{t }=v_{0}+at = 0+ 9.8*3.41 = 33.418 m/s

v_{t2 }- v_{02 }= 2as

depth of the well is s= (33.418^{2 }-0)/2/9.8 = 57 m

Given:

a = -9.8 m/s2

t = 3.41 s

vi = 0 m/s

Find:

d = ??

d = vi*t + 0.5*a*t2

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2

d = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)

d = -57.0 m

(NOTE: the - sign indicates direction)