Physics Question EXPERTS ONLY!!!! PROS!!!?

Question

Hint: First show that the first sound heard by the observer came from the jet when the line connecting the observer and the jet is perpendicular to its path. a jet flies toward higher altitude at a constant speed of 160m/s in a direction making an angle x with the horizontal. An observer (at x), on the ground, hears the jet for the first time when it is directly overhead. Determine the value of x if the speed of sound in air is Vsound = 339m/s. Answer in units of deg.

Answer 1

[Edit: Andy wrote, "The first sound will occur when the jet is the closest to the observer. This path must be perpendicular because the shortest distance between a line and a point is one that is perpendicular. "

It's not that simple. Before the jet reaches the perpendicular position, it emits sound, and that sound gets a head start. So there's a time lag that must be taken into account. Am still working the problem. End edit.]

x is an angle expressed in degrees. It doesn't make sense to me that the observer is located at x on the ground. I think that's a typo. Therefore I'll assume the observer is at the origin O.

We know that the jet passes directly overhead, so there's a vertical plane containing the jet's line of flight and the observer's position at O. That makes this a two-dimensional rather than a three-dimensional problem. Also, we'll ignore the speed of light

In the xy plane, draw a line representing the flight path with a positive slope m = tan x crossing the vertical y-axis at B. Draw the perpendicular from O, intersecting that line at A. Then OAB is a right triangle.

I'll do this problem backward, getting the answer first, and (maybe) later showing that the first sound came when the plane's path was perpendicular.

Assume that the first sound heard was emitted when the plane was at A. It took t seconds to reach the observer, with sound traveling at 339 m/s, so the distance AO = 339t.

In the same t seconds, the jet moved from A to B at 160 m/s, so AB = 160t.

By the geometry of the triangle OAB, angle AOB equals x, the angle of the jet's ascent. So tan x = AB/AO = (160t)/(339t) = 0.47198, and x = 25.27 degrees. (ANSWER)

Now the other part of this question is to show that the first sound reaching the observer was emitted when the jet was at A.

Place another point C on the jet's path to the left of A, so that angle COA = z < x. We need to show that sound emitted from A reaches the observer before sound emitted from C.

Triangle CAO is a right triangle; CO = AO/cos z; AC = AO tan z. A sound emits from point C, and it's heard at O at time

t1 = CO/339 = AO/(339 cos z)

The jet flies from C to A and emits another sound. That's heard at O at time

t2 = AC/160 + AO/339 = (AO/160) tan z + AO/339

We need to show that t2 < t1:

(AO/160) tan z + AO/339 < AO/(339 cos z)

(tan z)/160 < (1/cos z - 1)/339

(tan z)/(1/cos z - 1) < 160/339

(sin z)/(1 - cos z) < 160/339

So your conjecture -- that the first sound heard is emitted when the jet is at A (the perpendicular) -- depends on the above relationship for small z (z
Suppose z approaches zero (a very small angle, just before the perpendicular). Since sin 0 / (1 - cos 0) is indeterminate, let's try

lim (z ==> 0)[sin z / (1 - cos z) = lim (cos z / sin z) > a for any a.

That means the conjecture is wrong. The observer hears the sound from the jet the jet reaches the perpendicular at point A.

Maybe I made a mistake here. (I'm not going to check my work.) The point, though, is that this type of analysis must be done to solve this problem. It is not enough to say, as Andy has done, that the shortest distance between a point and a line is the perpendicular.

By the way, I have a hunch that your conjecture may be true if the jet is flying at supersonic speed. Just a hunch, though.

Answer 2

Call point 1 the place where the jet is when it emits the sound that the observer hears, point 2 is where the jet is when it is directly overhead. The first sound will occur when the jet is the closest to the observer. This path must be perpendicular because the shortest distance between a line and a point is one that is perpendicular. Now it just becomes a trig problem. Call d_p the distance the plane has to travel from point 1 to 2, d_o the distance along the perpendicular from the observer to point 1, and h is is the distance from the observer to point 2 directly overhead:

sin(x) = d_p/h
h = d_p/sin(x)
cos(x) = d_o/h
h = d_o/cos(x)
d_o/cos(x) = d_p/sin(x)
sin(x)/cos(x) = d_p/d_o
tan(x) = d_p/d_o

v = d/t
d = v*t
d_p = 160t
d_o = 339t

tan(x) = 160t/339t
tan(x) = 160/339
x = 25.27 deg

Answer 3

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