Calculating Sonic Boom Time: Jet Aircraft at Mach 1.5 and Altitude of 5029m

[inflames829]

Homework Statement

A jet aircraft traveling with a velocity of mach 1.5 at an altitude of 5029m passes directly over an observer on the ground. How many seconds elapse before the observer experiences the sonic boom created by the jet? speed of sound= 343ms-1

Homework Equations

The Attempt at a Solution

the only example i can find of this in my book has only this equation (its for the angle of the shock cone) sin theta = v sound/ v object. Can anyone give me any help on how i can work out the time from this, I am pretty sure you have to use this equation at some point. Can anyone give any help please.

 

[Dick]

Draw a picture. The sound will reach you when the shock cone hits you. It's a trig problem. Work out the horizontal distance the plane needs to travel for that. Then use that to find the time.

 

[inflames829]

i did that an got the angle as 0.8867. then i used tan-1 x adjacent equals opposite, which i had as the horizontal distance. Then i used time = distance/velocity but still get the wrong answer. Can you please help, am i still doing something wrong

 

[Dick]

vsound/vobject=1/1.5. I don't get sin(theta)=1/1.5 giving me theta=0.8867. Can you check that? And tan(theta)=height/distance traveled, right?

 

[inflames829]

i converted mach to meters a second but if i use 1/1.5 i get 0.73 degrees for theta. Using this i got 3171.18m for the horizontal distance. Now do i use time = distance/velocity, and if so do i keep the speed in mach or meters a second

 

[Dick]

I'll take 0.73 radians (NOT DEGREES) as a correct answer. But I still don't get your horizonal distance. Keep the velocity in m/sec. Mach is a ratio, you can't divide distance by mach and get seconds.

 

[inflames829]

i still get the same horizontal distance. I use tan theta = opposite/adjacent, which i rearrange to tan theta x adjacent = opposite, so tan 0.73 x 5029 = 4500.54. Is that right because i still get the wrong answer

 

[Dick]

Opposite=5029m, right? It's the height. distance=5029m/tan(theta). theta is the opening angle of the cone, innit?

 

[inflames829]

ohh, i didnt pick up on the rearranging error. Now I get the horizontal distance to be 5619.51223. Is that right. Then if i divide that by 442.5696 (which is mach 1.5) i get 12.70, but the correct answer is 11, am i still doing something wrong? do i use taht equation?

 

[Dick]

ohh, i didnt pick up on the rearranging error. Now I get the horizontal distance to be 5619.51223. Is that right. Then if i divide that by 442.5696 (which is mach 1.5) i get 12.70, but the correct answer is 11, am i still doing something wrong? do i use taht equation?

Have you thought about taking your calculator in for a tune-up? You never seem to get the same numbers I do. Mach 1.5 is (343m/sec)*1.5. I get 514.5m/sec, not 442.

 

[inflames829]

thanks. I got the right answer...in the end. I was using a converter for the mach to meters a second conversation that's why i got it wrong. But thanks for the help