Solved Problem on One-dimensional Motion
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A rocket is launched vertically from the ground at an initial speed of 200 m/s, after 10 seconds it explodes. A ground observer is located at a distance of 2000 meters in the same horizontal as the launch point. After how long he will hear the blast noise? The speed of sound in the air is 340 m/s.


Problem Data:
  • Initial speed of rocket:    v0 = 200 m/s;
  • Time to explosion:    t = 10 s;
  • Distance from observer to launch point:    d = 2000 m;
  • Acceleration due to gravity:    g = 9,8 m/s2;
  • Speed of sound in air:    vs = 340 m/s.
Problem diagram:

The rocket rises to a height h where it explodes, the sound of the blast traveling through the air for a time interval ts until it reaches a ground observer (Figure 1).

Figure 1

Solution

First, we must find the height reached by rocket. We choose a frame of reference in an upward direction (Figure 2).
The motion of the rocket is a vertical lauch, the height h of the explosion will be given by
\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_0+v_0 t-\frac{g}{2}\;t^2} \end{gather} \]
where the acceleration due to gravity has a negative sign because it has the opposite direction of the frame of reference, in the equation S=h and S0}=0
Figure 2
\[ \begin{gather} h=0+\left(200\;\mathrm{\small{\frac{m}{\cancel s}}}\right)\left(10\;\mathrm{\cancel s}\right)-\left(\frac{9.8\;\mathrm{\frac{m}{s^2}}}{2}\right)(10\;\mathrm s)^2\\[5pt] h=2000\;\mathrm m-490\;\mathrm m\\[5pt] h=1510\;\mathrm m \end{gather} \]
The Pythagorean Theorem gives the distance H from the point of the explosion to the observer (Figure 3)
\[ \begin{gather} H^2=(1470\;\mathrm m)^2+(2000\;\mathrm m)^2\\[5pt] H^2=2160900\;\mathrm{m^2}+4000000\;\mathrm{m^2}\\[5pt] H=\sqrt{6160900\;\mathrm{m^2}\;}\\[5pt] H\approx 2482\;\mathrm m \end{gather} \]
Figure 3

The gravitational field does not act on the sound wave, it moves at a constant speed. We choose a frame of reference with the origin at the point where the explosion occurs and oriented in the direction of the man (Figure 5).

Figure 5

Since S = 2482 m e S0 = 0, the equation of displacement as a function of time will be given by
\[ \begin{gather} S=S_0+v_{\small S} t_{\small S}\\[5pt] 2482\;\mathrm m=0+\left(340\;\mathrm{\small{\frac{m}{s}}}\right)t_{\small S}\\[5pt] t_{\small S}=\frac{2482\;\mathrm{\cancel m}}{340\;\mathrm{\frac{\cancel m}{s}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {t_{\small S}\approx 7.3\;\mathrm s} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .