Solved Problem on Acoustics

A projectile is fired at an angle of 60° to the horizontal and explodes when it hits the ground. Determine the initial velocity of the projectile and the range of the shot, knowing that the starting and ending points are in the same horizontal plane and that the air temperature is 25 °C, assume the speed of sound in air at 0 °C is equal to at 332 m/s.

Problem data:

Shooting angle: α = 60°;
Time interval between trigger and explosion: t = 18 s;
Air temperature: θ = 25 °C;
Speed of sound at 0 °C: v_S = 332 m/s;
Acceleration due to gravity: g = 9.8 m/s².

Problem diagram:

Figura 1

Solution

The speed of sound of the explosion, v_E, in the air at a temperature θ known at the speed at 0 °C, is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {v_{\theta }=v_{0}\sqrt{1+\frac{\theta }{273}}} \end{gather} \]

\[ \begin{gather} v_{E}=v_{S}\sqrt{1+\frac{\theta}{273}}\\[5pt] v_{E}=332\times\sqrt{1+\frac{25}{273}}\\[5pt] v_{E}=346.8\;\text{m/s} \end{gather} \]

since the speed of sound of the explosion is in the opposite direction to the direction of the trajectory, it will have a negative sign

\[ \begin{gather} v_{E}=-346.8\;\mathrm{m/s} \tag{I} \end{gather} \]

The time interval between the shot and the explosion will be the time that the projectile takes to travel the path, t_P, added to the time that the sound of the explosion takes to return, t_E, this time interval is given equals to 18 s

\[ \begin{gather} t=t_{P}+t_{E}\\[5pt] t_{P}+t_{E}=18 \tag{II} \end{gather} \]

We decompose the movement in the directions of the x and y axes (Figure 2).
Along the x-axis, we have a motion with constant speed and no acceleration acting in this direction. If we take equal time intervals, then the intervals (Δx₁, Δx₂, Δx₃, Δx₄, Δx₅, Δx₆) of space covered will all be equal. The sound from the explosion will also travel along the x-axis with constant velocity from D to origin 0.

Figure 2

Along the y-axis, we have a motion at constant acceleration, the acceleration due to gravity is acting in this direction, and we have a vertical launch motion on the way up and a free fall on the way down. If we take equal time intervals, we will have during the ascent of the projectile that the intervals traversed will be smaller and smaller (Δy₁ > Δy₂ > Δy₃), the acceleration due to gravity acts in the downward direction decreasing the speed of the projectile while it rises. When the projectile is descending, the acceleration due to gravity increases the speed of the projectile, so the displacements traveled will be greater and greater (Δy₄ < Δy₅ < Δy6).

The projectile velocity, which is given by making an angle of 60° with the x-axis, can also be projected along the x and y-axes (Figure 3), the initial velocity of the projectile has the components

\[ \begin{gather} v_{0x}=v_{0}\cos \alpha \tag{III}\\[8pt] v_{0y}=v_{0}\sin \alpha \tag{IV} \end{gather} \]

The equation of motion along the x-axis will be

\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+vt} \end{gather} \]

Figure 3

\[ \begin{gather} S_{x}=S_{0x}+v_{0x}t \end{gather} \]

as the projectile starts from the origin S_0x = 0, t = t_P is the time interval that the projectile takes to travel the path to the target, from 0 to D, v_0x is given by the expression (III) and α = 60º, substituting

\[ \begin{gather} S_{Px}=v_{0}\cos 60°t_{P} \end{gather} \]

From Trigonometry \( \cos 60°=\dfrac{1}{2} \)

\[ \begin{gather} S_{Px}=\frac{1}{2}v_{0}t_{P} \tag{V} \end{gather} \]

The equation of motion along the y-axis will be

\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\frac{a}{2}t^{2}} \end{gather} \]

\[ \begin{gather} S_{y}=S_{0y}+v_{0y}t-\frac{g}{2}t^{2} \end{gather} \]

The projectile starts from the origin S_0y = 0, t = t_P is the time interval to follow the path (up and down), v_0y is given by the expression (IV), and the acceleration a = g, as the acceleration due to gravity is in the opposite direction to the reference frame, it will be negative

\[ \begin{gather} S_{Py}=v_{0}\sin \alpha t_{P}-\frac{9.8}{2}t_{P}^{2} \end{gather} \]

From Trigonometry \( \sin 60°=\dfrac{\sqrt{3\;}}{2} \)

\[ \begin{gather} S_{Py}=\frac{\sqrt{3}}{2}v_{0}t_{P}-4.9t_{P}^{2} \tag{VI} \end{gather} \]

For the motion in the y direction, the equation for the velocity of the projectile is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {v=v_{0}+at} \end{gather} \]

\[ \begin{gather} v_{y}=v_{0y}-gt \end{gather} \]

where v_0y is given by the expression (IV) and the acceleration a = g

\[ \begin{gather} v_{Py}=v_{0}\sin \alpha-9.8t_{P}\\[5pt] v_{Py}=\frac{\sqrt{3\;}}{2}v_{0}-9.8t_{P} \tag{VII} \end{gather} \]

The time interval that the projectile will take to reach the maximum height is calculated using equation (VII), when the velocity component in the y direction is zero, in this direction the projectile stops and its velocity reverses direction (the component in the x direction does not change, it keeps going forward). If t_SP is the time of the rise of the projectile and letting v_Py = 0

\[ \begin{gather} 0=\frac{\sqrt{3\;}}{2}v_{0}-9.8t_{P}\\[5pt] 9.8t_{P}=\frac{\sqrt{3\;}}{2}v_{0}\\[5pt] t_{P}=\frac{\sqrt{3\;}}{2\times 9.8}v_{0} \tag{VIII} \end{gather} \]

The total time for the projectile to travel the path will be the ascent time added to the descent time, as these are equal, the total time will be twice the value found in (VIII)

\[ \begin{gather} t_{P}=2\times\frac{\sqrt{3\;}}{2\times 9.8}v_{0}\\[5pt] t_{P}=\frac{\sqrt{3\;}}{9.8}v_{0} \tag{IX} \end{gather} \]

The range of the projectile can be calculated with equation (V), where S_Px = D and the time interval to cover the entire path will be the value calculated above in expression (IX)

\[ \begin{gather} D=\frac{1}{2}v_{0}\times\frac{\sqrt{3\;}}{9.8}v_{0}\\[5pt] D=\frac{\sqrt{3\;}}{19.6}v_{0}^{2} \tag{X} \end{gather} \]

Sound propagation also has a motion with a constant speed, for the sound, the initial position will be the explosion point S_0E = D, and for the initial velocity, the value found in (I) will be used.

\[ \begin{gather} S_{E}=S_{0E}+v_{0E}t_{E}\\[5pt] S_{E}=D-346.8t_{E} \tag{XI} \end{gather} \]

the final position will be the origin, from where the shot was fired, equal to zero, S_E = 0, and the time interval can be obtained from equation (II), t_E = 18 −t_p

\[ \begin{gather} 0=D-346.8\times(18-t_{P}) \end{gather} \]

for D, we use the value of (X), and for t_P, we substitute the value of (IX)

\[ \begin{gather} \frac{\sqrt{3\;}}{19.6}v_{0}^{2}-346.8\times\left(18-\frac{\sqrt{3\;}}{9.8}v_{0}\right)=0\\[5pt] \frac{\sqrt{3\;}}{19.6}v_{0}^{2}-346.8\times 18-346.8\times\frac{\sqrt{3\;}}{9.8}v_{0}=0\\[5pt] 0.088v_{0}^{2}+61.293v_{0}-6242.4=0 \end{gather} \]

This is a Quadratic Equation in v₀.

Solution of \( 0.088v_{0}^{2}+61.293v_{0}-6242.4=0 \)

\[ 0.088v_{0}^{2}+61.293v_{0}-6242.4=0 \]

\[ \begin{array}{l} \Delta=b^{2}-4ac=(61.293)^{2}-4\times 0.088\times(-6242.4)=5954.157\\[10pt] v_{0}=\dfrac{-b\pm\sqrt{\Delta \;}}{2a}=\dfrac{-61.293\pm\sqrt{5954.157\;}}{2\times 0.088}=\dfrac{-61.293\pm 77.163}{0.176} \end{array} \]

the roots will be

\[ \begin{gather} v_{01}=90.1\\[5pt] \text{ou}\\[5pt] v_{02}=-786.7 \end{gather} \]

since we chose that the cannon fires in the positive direction of the frame of reference, we neglected the negative root of the solution will be

\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{0}=90.1\;\text{m/s}} \end{gather} \]

Substituting this value into the expression (X) for the range

\[ \begin{gather} D=\frac{\sqrt{3\;}}{19.6}\times(90.1)^{2} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {D=717.4\;\text{m}} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .