Respuesta :

samuraijack
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

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