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A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.

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asked Jul 17, 2015 in Physics by carol (1,050 points)

3 Answers

0 votes

v0=0 m/s

t=3.41s

acceleration rate is a=g=9.8 m/s2

v=v0+at = 0+ 9.8*3.41 = 33.418 m/s

vt2 - v02 = 2as

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

answered Jul 18, 2015 by hx_l (660 points)
0 votes

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)

answered Jul 19, 2015 by dh (1,350 points)
0 votes
h=ut+1/2gt^2                      here,t=3.41s; u=0 ;g=9.8 m/s^2;h=?

=0+1/2*9.8*(3.41)^2    
=56.98 m

=57 m
answered May 26 by Borno
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