Question

a stone is dropped into a deep well and is heard to hit the water 3.41 seconds after being dropped. determine the depth of well ? (take g=9.8 m/s Square)

Answer 1

T = 3.41s

u = 0

S=ut+1/2*a*t^2

S=0(3.14)+1/2*9.8*(3.41)^2

S=0+4.9*11.62

S=56.93m

Answer 2

Answer:

The depth of the well is 48.31m.

Explanation:

We will use Newton's second equation of motion to solve this question,

$S=ut+\frac{1}{2} gt^2$       (1)

Where,

S=depth to which the body falls

u=initial velocity of the body

t=time taken

g=acceleration due to gravity=9.8m/s²

From the question we have,

u=0            

t=3.14 seconds

By substituting the value of u and t in equation (1) we get;

$S=0\times 3.14+\frac{1}{2}\times 9.8\times (3.14)^2$

$S=48.31m$

Hence, the depth of the well is 48.31m.