Doppler Effect/Projectile Motion

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Homework Statement

Find the frequency heard as you walk away from a building that has a fire alarm going off with a frequency of 2650 Hz. The velocity of the alarm is equivalent to the final velocity of an object being launched with an initial velocity of 235.4 km/h [W23°N] from a height of 5321 m. Assume air temperature to be -8.63°C.

This was a very confusing question on a recent assignment. This is exactly how it is worded.

The underlined parts make no sense to me,
"You walk away from a building" this would mean that v_obs has a value greater than 0, but this value is not given therefore I think the question is impossible.
"The velocity of the alarm" This makes no sense since the alarm should be fixed to the building. I assumed it to have a velocity anyways in my solution.

Homework Equations

Doppler Effect Equation:
f_obs = frequency detected by the observer
f₀ = Actual frequency
v_s = speed of sound
v_obs = velocity of the observer
v_sou = velocity of the source

f_obs = (v_s+v_obs/v_s+v_sou)f₀

For the projectile motion part

Trigonometric ratios to find the vertical and horizontal components of initial velocity.

v₂² = v₁² + 2aΔd

Pythagorean theorem to find final velocity using vertical and horizontal components.

The Attempt at a Solution

Projectile motion part

Givens:
v₁ = 235.4 km/h = 65.39 m/s [W23°N]
θ = 23°
d_y = 5321 m
a_y = 9.8 m/s²

Solution:
-Find vertical and horizontal component of initial velocity:

sinθv₁ = v_1y
25.55 m/s = v_1y

cosθv₁ = v_x
60.19 m/s = v_x

velocity on the x-axis is assumed to be constant.

-Find vertical component of final velocity:

v_2y² = v_1y² + 2ad
v_2y = 323.95 m/s

-Find final velocity

v₂² = v_2y² + v_x²
v₂ = 329.49 m/s



Doppler Effect Part

Givens:
f₀ = 2650 Hz
T = -8.63°C
v_sou = 329.49 m/s

-Find the speed of sound

v_s = 331.4 + (0.606)(-8.63)
v_s = 326.17 m/s

-Find the frequency heard by the observer

f_obs = (v_s + v_obs/v_s + v_sou)f₀

Missing v_obs. Is the question impossible?

 

[Brage]

By the velocity of the alarm they mean the velocity of the alarm with respec to you at a stand still. This is just a change of reference, it is the same as saying the velocity of you with respect to the alarm, or your speed if the alarm stood still (all be it in the oposite direction). i.e. speed of the alarm would then be the speed source-speed of the observer.