Free fall in a well

Data

t = 1.5 s
v_sound = 340 m/s
g = 9.8 m/s2
H = ?
 

Resolution

Let's consider the exercise in two parts:

• Part 1. The stone descends to impact the water at the bottom of the well
• Part 2. The sound moves up from the water to the opening of the well

If we call the time the stone takes to get to the water t₁ and t₂ the time the sound takes to get back to the opening of the well, we get that:

$$t=t_1+t_2$$

We will study each part separately:

Part 1

Applying the equation of position of the fall free motion and knowing that at t₁, the position of the stone is y = 0 m:

$$\begin{gathered} y=H-\frac{1}{2}·g·t^2 \Rightarrow \\ 0=H-0.5·9.8·{t_1}^2 \Rightarrow \\ \underline{H=4.9·{t_1}^2} \end{gathered}$$

Part 2

Sound rises at constant speed, and we assume that it does it in a straight line. Therefore:

$$\begin{gathered} x=v·t \Rightarrow \\ H=v_{sonido}·t_2 \Rightarrow \\ \underline{H=340·t_2} \end{gathered}$$

Since H is the same in both equations and we know that t=t₁+t₂:

$$\begin{gathered} \left.\begin{array}{r}4.9·{t_1}^2=340·t_2\\ 1.5=t_1+t_2\end{array}\right\} \Rightarrow \\ \left.\begin{array}{r}4.9·{t_1}^2-340·t_2=0\\ t_2=1.5-t_1\end{array}\right\} \Rightarrow \\ 4.9·{t_1}^2-340·\left(1.5·t_1\right)=0 \Rightarrow \\ 4.9·{t_1}^2-340·t_1-510=0 \Rightarrow \\ {t}_1=\frac{-340\pm \sqrt{{340}^2-1·4.9·(-510)}}{2·4.9} \Rightarrow \\ \underline{t_1=1.47 s} \end{gathered}$$

Substituting t₁ in the equation of part 1:

$$\begin{gathered} H=4.9·{\left(1.47\right)}^2 \Rightarrow \\ \boxed{H=10.59 m} \end{gathered}$$