Projectile motion - microgravity astronaut training

[nebullient]

Homework Statement

In microgravity astronaut training and equipment testing, NASA flies an aircraft along a parabolic flight path. The aircraft climbs from 24,000 ft to 31,000 ft, where it enters a parabolic path with a velocity of 143 m/s nose high at 45.0° and exits with velocity 143 m/s at 45.0° nose low. During this portion of the flight, the aircraft and objects inside its padded cabin are in free fall; astronauts and equipment float freely as if there were no gravity. What are the aircraft's a) speed and b) altitude at the top of the maneuver? c) What is the time interval spent in microgravity?

Homework Equations

The Attempt at a Solution

I'm stuck on part c. I'm using the equation d_y = -1/2gt² + v_iyt. I used 143sin45° for v_iy and 522 m for d_y, which was the answer for part b. However, when I solved this quadratic equation, I got no solution. (I would then multiply this by 2.)

I also attempted to use v_fy = v_iy + at, and multiplied by 2 to get 20.6 s, which is the right answer.

I'm really confused as to why the first method doesn't work, though.

 

[BvU]

Hello nebullient,

I used 143sin45° for viy

Free fall does not start with an initial vertical velocity (this value and orientation is achieved when back down to 24000 ft).

 

[BvU]

20.6 s, which is the right answer.

do you guys use 0.3 m/foot ?

 

[nebullient]

Hi, BvU! Thank you for the welcome :)

I used that value for part b though, and I got it right?

And yes, I did use 0.3 m/ft for part b.

 

[jbriggs444]

As described in this problem statement, free fall starts with a non-zero initial vertical velocity. A calculation that equates $d_y$ with 522 will be covering only half of the free fall interval, the part from entry to apex. Still, that should not account for a failure to have a solution for the quadratic.

One thing that comes to mind is that this quadratic is ill-conditioned. You are trying to find the place where the graph of a quadratic function (a parabola opening downward) just barely touches the horizontal line corresponding to a previously calculated peak altitude. By design, it should just touch.

A tiny bit extra vertical velocity and it will do more than touch. There will be two zeroes.
A tiny bit less vertical velocity and it won't touch. There will be no zeroes.

Any round-off error in your calculation and the discriminant in the quadratic formula can go negative. If you do the calculations exactly, it should work out that the discriminant in the quadratic formula is exactly zero. If it's significantly negative, you've messed something up. Show your work!

But, better yet, skip the ill conditioned quadratic formula and try another approach. Ask yourself how long it takes to decelerate to zero vertical velocity if starting at $v_{iy}$.

 

[nebullient]

@jbriggs444 Wow, it really was due to rounding! I carried it out to three decimal places and I got the correct answer. Thank you so much! I've been stuck on this for so long; this truly shows the importance of using exact values.

 

[jbriggs444]

@jbriggs444 Wow, it really was due to rounding! I carried it out to three decimal places and I got the correct answer. Thank you so much! I've been stuck on this for so long; this truly shows the importance of using exact values.

My "lesson learned" on such things is somewhat different.

Dealing with rounding errors by throwing more precision at the problem is not always the best way to proceed. One needs to be alert to the possibility of such problems (of course). But it is often better to switch formulas or approaches rather than to simply add extra digits and pray.

 

[nebullient]

I agree, which is why I used another formula (see the original post). I just wanted to know why this formula wasn't working, because it should have.

 

[BvU]

Free fall does not start with an initial vertical velocity (this value and orientation is achieved when back down to 24000 ft).

Have to apologize for wrongfooting: Blindly accepting the factor 2, I went from 31000 ft horizontal to 24000 ft and used $\ \Delta h = {1\over 2} gt^2\$ yielding 20.48 s. (1 ft = 12", 20.7 s for 1 ft = 0.3 m; g = 9.81 m/s²)

And for a stupid remark . Free fall starts when engine acceleration stops. The initial vertical velocity at 24000 ft is of course the 143 $\sin 45^\circ$ you mention.

 

[jbriggs444]

And for a stupid remark . Free fall starts when engine acceleration stops. The initial vertical velocity at 24000 ft is of course the 143 $\sin 45^\circ$ you mention.

I think that's still a slightly wrong picture. There is a better one here.