Physics, asked by georgekbinoj2003, 11 months ago

An airplane is in supersonic flight at an altitude h. At what smallest distance a (along the horizontal) from the
observer on the ground is there a point from which the sound emitted by the airplane motors travels to the
observer faster than from point A that is directly above the observer

Answers

Answered by jhumakhamrai097
3

Answer:

t = √(a^2+h^2 )/v_s => t_2=a/v_p +h/v_s ; t1 < t2¬

a < 2(v_p/v_s )/((v_p/v_s )^2-1) h

Answered by vaibhavsemwal
1

Answer:

Required distance = a &lt; \frac{2v_s v_p h}{v_p^2-v_s^2}

Explanation:

Airplane at at an altitude = height = h

Say, airplane is at a horizontal distance a from the observer.

Now, Distance between airplane and observer =\sqrt {a^2+h^2}

Let, v_p be the speed of plane.

and v_s be the speed of sound.

Time taken by sound to reach observer intially = t_1 = \frac{\sqrt {a^2+h^2}}{v_s}

And, when airplane has reached overhead, then time,

t_2 = \frac{a}{v_p} +\frac{h}{v_s}

As,

t_1 &lt; t_2

\implies \frac{\sqrt {a^2+h^2}}{v_s} &lt; \frac{a}{v_p} +\frac{h}{v_s}

\implies a &lt; \frac{2v_s v_p h}{v_p^2-v_s^2}

\implies a &lt; 2(v_p/v_s)h/((v_p/v_s)^2-1)

#SPJ2

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