Question

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.

Answer 1

Explanation:

Given, acceleration of stone is a=g=9.8m/s

2

Initial velocity is u=0m/s

Time for hitting water is t=3.41s

If d be the depth of the well.

Using d=ut+at

2

/2,

∴ d=0(3.41)+0.5(9.8)(3.41)

2

=56.98∼57m

Answer 2

Answer:

• Depth of well = 56.98 metres  [approx.]

Explanation:

Given that,

Stone is dropped into a deep well, and is heard to hit the water after 3.41 s of being dropped.

so,

• Initial velocity of stone, u = 0
• acceleration due to gravity, a = 9.8 m/s²
• time taken by stone to reach the water surface, t = 3.41 s

We need to find,

Depth of well

that is,

• Distance covered by stone, s =?

Formula required,

• Second equation of motion

       s = u t + 1/2 a t²

[ Where s = distance covered; u = initial velocity, a = acceleration; t = time taken ]

Calculation,

Using second equation of motion

→ s = u t + 1/2 a t²

→ s = ( 0 ) ( 3.41 ) + 1/2 ( 9.8 ) ( 3.41 )²

→ s = ( 4.9 ) ( 3.41 ) ( 3.41 )

→ s = 56.98  m  [approx.]

Therefore,

• Depth of the well is 56.98 metres approximately.