A stone is dropped into a well. the splash is heard 1.73 s later. what is the depth of the well?

In this problem the only given is the time it take the stone in reaching the surface water on the well which is t=1.73s. Remember the action of the stone is going downward, therefore the only force that act on the stone is the normal pull of gravity "g" which is equivalent to 9.8m/s². The question is the distance going down or the depth of the well. Simply follow this formula. d=1/2gt²

distance=multiply "g" to the square of time and divided by two.

Let x = the depth of the well

Let t₁ = the time of travel to reach the water surface in the well.

Then

x = (1/2) * g*t₁² = 0.5 * (9.8 m/s²) * (t₁ s) ² = 0.49t₁².

t₁ = √ (x/0.49) = 1.4286√x s

The speed of sound is 343 m/s.

Let t₂ = the time for the sound of the impact to travel x m.

Then

t₂ = (x m) / (343 m/s) = 0.002915x s.

The time that elapsed before the impact was heard is 1.73 s.

Therefore

t₁ + t₂ = 1.73

1.4286√x + 0.002915x = 1.73

1.4286√x = 1.73 - 0.002915x

2.0949x = 2.9929 - (1.00859 x 10 ⁻²) x + (8.4972 x 10⁻⁶) x²

(8.4972 x 10⁻⁶) x² - 2.105x + 2.9929 = 0

x² - (2.4773 x 10⁵) x + 3.5222 x 10⁵ = 0

Solve with the quadratic formula.

x = 0.5[2.4773 x 10⁵ + / - √ (6.1369 x 10¹⁰) ]

= 2.477 x 10⁵ m, or 1.422 m

Because (2.477 x 10⁵ m) / (343 m/s) = 722 s (greater than 1.73 s), this solution is rejected.

Therefore

x = 1.422 m

Answer: 1.422 m