Question

b) A stone is dropped
into a well. The Sound
the splash is heard
4'0 second later. What
the depth
depth of
the well a
tu
is​

Answer 1

Explanation:

assumptions,

Let H denotes the depth (in m) of the well, which is assumed to be the height of open end at the ground level from the water level in the well.

Let T1 denotes the time (in s) of free fall of the stone in the well from the open end at the ground level to the water level in the well.

Let T2 denotes the time (in s) required for the sound of splash to travel from the water level in the well to its open end at the ground level.

Let V denotes velocity (in m/s) of the stone when it reaches the water level in the well after it is dropped from open end at the ground level.

Let M denotes the mass (in kg) of the stone.

Hence from above data we get following relations,

T1 + T2 = 4 …. (1a)

V = 0 + g*T1 [g = gravitational acceleration]

or V = g*T1 …. (1b)

350*T2 = H

or T2 = H/350 …. (1c)

M*g*H = (1/2)*M*V^2 [P.E. of the stone at the ground level = K.E. of the stone at water surface level]

or H = (1/2)*V^2/g

or H = (1/2)*(g*T1)^2/g [from (1b)]

or H = (1/2)*g*T1^2

or H = (1/2)*g*(4 - H/350)^2 [from (1a) & (1c)]

or H = (1/2)*9.8*(4 - H/350)^2 [g = 9.8 m/s^2 (assumed)]

or 350*h = 4.9*(4 - h)^2 [h = H/350 (assumed for simplification purpose)]

or 350*h = 78.4 - 39.2*h + 4.9*h^2

or 4.9*h^2 - 389.2*h + 78.4 = 0

or h = [389.2 ± √(389.2^2 - 4*4.9*78.4)]/9.8

or h = (389.2 ± 387.2)/9.8

or h = 0.204 [the other value h=79.224 is rejected because it generates unreasonable absurd value for H (27728.4 m)]

Therefore the required depth of the well,

from the ground level to the water surface level is given by,

H = 350*h = 350*0.204 = 71.4 (m) [Ans]