Sonic boom trig question

Question

The US Air Force was running maneuvers with the F22 Raptor, flying low enough so that you could see the shock cone created as the jet flies overhead at supersonic speeds. You wanted to take a video of this. You stand on the ground and estimate you are 500 feet below the plane as it passes overhead from left to right. You wish to express distance d from your camera to the jet as a function of the angle x. Consider the domain of the function to be             -pi/2<x<pi/2

 

a. Draw a picture to model the situation 

 

b. Thinking about the scenario, which trigonometric function should be used to model this data and explain why

 

c. Write the equation to model this scenario. Explain why the given domain makes sense

 

d. Determine the distance of the jet from the camera for angles of -60 degrees, -45 degrees, 0 degrees, and 30 degrees. What do the distances mean in context of the problem (you should be thinking 'from what direction')

Answer 1

Go ahead and draw a picture of this situation.

 

At any given time, you'll have a triangle with vertices of yourself, the plane, and a spot in the air above you 500 feet up. The plane will appear to trace an arc overhead, starting at one horizon (at an apparent height of 0), passing through the point in the air above you, and ending at the other horizon (again at an apparent height of 0).

 

The angle x is located where you stand. When the plane appears on the horizon, x is an angle of 90 degrees (π/2). As the plane grows nearer, the angle shrinks until it reaches 0 degrees when the plane is above you. Then, the angle widens again as the plane grows farther away (although the angle will be pointing in the opposite direction) until it again reaches 90 degrees. Call angles facing towards the plane's appearance (left) negative; call angles facing towards the plane's disappearance (right) positive. So, the angle x starts at -90 degrees (-π/2) and passes through 0 to 90 degrees (π/2) as the plane goes by.

 

 

 

We're interested in the distance from camera to jet. The angle located in the air above you is always a right angle, since it's directly above you and level with the plane. So, the distance from camera to plane is the hypotenuse (across from the right angle). You also know the distance from you to the point in the air (500 feet), which is the side adjacent to angle x. So, you'll want to use cosine, which relates adjacent side to hypotenuse.

 

cos(x) = adjacent/hypotenuse

cos(x) = 500/d

d(cos(x)) = 500

d = 500/cos(x)

 

 

 

d = 500/cos(-60)

d = 500/(1/2)

d = 1000 feet

 

When the plane is 1/3 of the way from the horizon to your left to directly above you, it's 1000 feet away.

 

d = 500/cos(-45)

d = 500/(√(2)/2)

d = 1000/√(2)

d = 1000√(2)/2

d = 500√(2) feet

 

When the plane is 1/2 of the way from the horizon to your left to directly above you, it's 500√(2) feet away.

 

d = 500/cos(0)

d = 500/1

d = 500 feet

 

When the plane is directly above you, it's 500 feet away.

 

d = 500/cos(30)

d = 500/(√(3)/2)

d = 1000/√(3)

d = 1000√(3)/3

 

When the plane is 1/3 of the way from directly above you to the horizon to your right, it's (1000√(3))/3 feet away.