"Vomit Comet" SpaceCraft (0)-V & H at Max Height

[Const@ntine]

Homework Statement

Physics for Scientists and Engineers, 8th Edition, Ex. 47 M4

A spacecraft flies from 24.000 ft to 31.000 ft, in which it enters a parabolic trajectory with a velocity of 143 m/s at an angle of 45, above the horizontal. It exits with a velocity of 143 m/s at an angle of 45 bellow the horizontal. In that part of its flight, the spacecraft is in a "free fall" ; the astronauts and equipment fly as if there's no gravity.

a) What's the magnitude of the Velocity [...]
b) [...] and the height of the spacecraft at the maximum height of the maneuver?
c) For how much time do the zero g conditions apply?

Translation:

Left Side: Height (ft)

Bottom Horizontal: Time of Maneuver (s)

First Part: Lifting of snoot at 45

Second Part: Zero g

Third Part: Lowring of snoot at 45

Homework Equations

Yf=Yi + Viy*t + 1/2*a*t^2
Vf=Vi + a*t
Xf=Xi + a*t

The Attempt at a Solution

I'm going to be honest here, I have no clue what to do. At first I thought that it started at 24k ft with 143 m/s, so I set out to find his Velocity at 31k ft, but obviously that was wrong. The I figured that since there's no gravity while it's in the parabolic trajectory, it'd have a constant speed of V = Vi*sin(45) = 101,1 m/s which is close enough to the book's answer (101 m/s).

Thing is, I can't for the life of me get how in zero g environment, a spacecraft could reach a bigger height and then start falling again. I'm obviously missing some key component, so I'd greatly appreciate any kind of help.

 

[PeroK]

I think it's the part of the flight above 31,000 ft that is free fall: only the middle bit. Very confusing.

 

[TomHart]

Can't this problem just be thought of as shooting a bullet at an initial velocity of 143 m/s at an angle of 45 degrees at an initial height of 31,000 feet? Okay, granted, it's a fairly slow bullet. So no, it will not have a constant velocity in that free fall portion.

Would you know how to work that bullet problem?

 

[Const@ntine]

Can't this problem just be thought of as shooting a bullet at an initial velocity of 143 m/s at an angle of 45 degrees at an initial height of 31,000 feet? Okay, granted, it's a fairly slow bullet. So no, it will not have a constant velocity in that free fall portion.

Would you know how to work that bullet problem?

A bullet being shot from a gun with Vi=143 m/s at a 45 angle with g, right? Yeah, I could do that, but the problem's statement says that g is zero in that particular part. Mabe it's a translation problem, since mine is not the original version in english.

 

[PeroK]

A bullet being shot from a gun with Vi=143 m/s at a 45 angle with g, right? Yeah, I could do that, but the problem's statement says that g is zero in that particular part. Mabe it's a translation problem, since mine is not the original version in english.

Free fall is zero gravity. It means zero gravity inside the plane, as everything inside is in free fall along with the plane.

 

[TomHart]

Well, a bullet would experience 0g in that particular portion of flight. It would be in a free fall. When the problem says "zero g", it actually means that it is accelerating downward at 1 g. I know that sounds strange. But here's how I look at it. Think of it as the pilot sitting on a scale - or better yet, a scale that can measure in any direction. When he is in a free fall, the scale would weigh the pilot to be 0 pounds. That's what they mean by zero g. So if a pilot was flying horizontally at constant speed, that would be considered 1 g in flight terminology. At least I think that is how it works. There are probably some pilots on here that could confirm or deny that - and also elaborate.

Edit: I thought the translation was pretty good.

 

[Const@ntine]

Free fall is zero gravity. It means zero gravity inside the plane, as everything inside is in free fall along with the plane.

Well, a bullet would experience 0g in that particular portion of flight. It would be in a free fall. When the problem says "zero g", it actually means that it is accelerating downward at 1 g. I know that sounds strange. But here's how I look at it. Think of it as the pilot sitting on a scale - or better yet, a scale that can measure in any direction. When he is in a free fall, the scale would weigh the pilot to be 0 pounds. That's what they mean by zero g. So if a pilot was flying horizontally at constant speed, that would be considered 1 g in flight terminology. At least I think that is how it works. There are probably some pilots on here that could confirm or deny that - and also elaborate.

Edit: I thought the translation was pretty good.

Well, I tried to see it as a usual problem of "we throw something at an angle", and while some results match up, I've got a question:

-It says that it enters with 143 m/s at a 45 angle, right? And then it does a free fall, so his velocity goes from that number to zero (since a = -g = -9.8 m/s^2), and then the spacecraft starts accelerating again so that it exits with the same velocity. So, what does the statement mean by "magnitude of Velocity and Height at the maximux height of the maneuver" ? Isn't that when it technically becomes zero momentarily? The book says that it's 101 m/s, but that the velocity with which it enters the parabolic, at 31k ft. The maximum height I find is pretty close to the one in the book, and the time of the maneuvers match up, but question a is a problem.

 

[PeroK]

Well, I tried to see it as a usual problem of "we throw something at an angle", and while some results match up, I've got a question:

-It says that it enters with 143 m/s at a 45 angle, right? And then it does a free fall, so his velocity goes from that number to zero (since a = -g = -9.8 m/s^2), and then the spacecraft starts accelerating again so that it exits with the same velocity. So, what does the statement mean by "magnitude of Velocity and Height at the maximux height of the maneuver" ? Isn't that when it technically becomes zero momentarily? The book says that it's 101 m/s, but that the velocity with which it enters the parabolic, at 31k ft. The maximum height I find is pretty close to the one in the book, and the time of the maneuvers match up, but question a is a problem.

Why would velocity be $0$ at the highest point?

 

[TomHart]

The vertical component of velocity becomes 0 at the maximum height, but the horizontal component does not change. It remains constant ALWAYS. Also, the spacecraft (or plane, or bullet) does not "start accelerating again". Once it starts into that parabolic portion of flight, it is ALWAYS accelerating at 1 g in the downward direction.

 

[Const@ntine]

Why would velocity be $0$ at the highest point?

The vertical component of velocity becomes 0 at the maximum height, but the horizontal component does not change. It remains constant ALWAYS. Also, the spacecraft (or plane, or bullet) does not "start accelerating again". Once it starts into that parabolic portion of flight, it is ALWAYS accelerating at 1 g in the downward direction.

Yeah, I get it now. His vertical velocity becomes momentarily zero at the highest point, while his vertical velocity is constant, thus the magnitude of his Velocity there is 101,1 m/s. It's constantly accelerating (poor choice of words there, I know that g doesn't become zero when say, we're throwing a rock upwards, Vy becomes zero at some point, not the acceleration), it's just that after it reaches the highest point of the manuever, his vertical velocity starts increasing, in order for it to have a 143 m/s magnitude when it exits the parabolical trajectory, at 45, but bellow the horizontal this time.

 

[PeroK]

Yeah, I get it now. His vertical velocity becomes momentarily zero at the highest point, while his vertical velocity is constant, thus the magnitude of his Velocity there is 101,1 m/s. It's constantly accelerating (poor choice of words there, I know that g doesn't become zero when say, we're throwing a rock upwards, Vy becomes zero at some point, not the acceleration), it's just that after it reaches the highest point of the manuever, his vertical velocity starts increasing, in order for it to have a 143 m/s magnitude when it exits the parabolical trajectory, at 45, but bellow the horizontal this time.

That's correct. One thing to learn here is that the question confused you by putting a standard projectile motion problem in a new disguise. It's important to learn that not everything in a question may be relevant and to see a simpler problem even when it's disguised as something more complicated.

 

[Const@ntine]

That's correct. One thing to learn here is that the question confused you by putting a standard projectile motion problem in a new disguise. It's important to learn that not everything in a question may be relevant and to see a simpler problem even when it's disguised as something more complicated.

Yeah, I've got to work on that a bit. It's just that time is running out and the syllabus is huuuuuuuuge, so I'm kinda on the edge right now.

Either way, thanks for the help both of you!