Answer:
(a) Angle is 42.865°
(b) t=12.54s
Explanation:
The jet plane is at 5870m from Earth surface
The jet traveling at speed of 1.47 Mach
Speed of Sound v=343 m/s
Part (a)
As we know :
[tex]Sin\alpha =\frac{v}{vs}\\[/tex]
Substitute the given values
[tex]Sin\alpha =\frac{v}{1.47v}\\ Sin\alpha =\frac{343m/s}{1.47(343m/s)} \\Sin\alpha =0.68027\\\alpha =Sin^{-1}(0.68027)\\\alpha =42.865^{o}[/tex]
Angle is 42.865°
Part (b)
As the distance given as:
[tex]d=vst[/tex]
By using some trigonometry
[tex]d=\frac{h}{tan\alpha }[/tex]
Where h is the height of jet plane
Thus
[tex]t=\frac{h}{tan\alpha } \frac{1}{vs}[/tex]
Substitute the given values
So
[tex]t=\frac{5870m}{tan(42.865)}\frac{1}{1.47(343m/s)} \\t=12.54s[/tex]
