NASA Free-Fall Research: Solving Problems with Kinematic Equations

  • Thread starter Thread starter intriqet
  • Start date Start date
AI Thread Summary
The discussion revolves around two physics problems related to free-fall research at NASA. In the first problem, the participant correctly calculates the maximum time for free fall and the distance fallen in 5.24 seconds, but struggles with determining the required constant acceleration to stop the package in the remaining distance of 11 meters. In the second problem, the participant accurately finds the time for two stones to hit the water and the speeds at impact, but is unable to determine the necessary initial velocity for the second stone to hit simultaneously with the first. The participant notes that their calculations for acceleration and initial velocity seem straightforward but yield incorrect results in the Webassign system. The discussion highlights challenges in applying kinematic equations to real-world scenarios.
intriqet
Messages
21
Reaction score
0
I have two problems. I have filled in most of the information that is asked but I can't seem to figure out a part for each problem. Please Help.

Problem 1:

Homework Statement



At a NASA research center free-fall research is performed by dropping experiment packages from the top of an evacuated shaft 146 m high. Free-fall imitates the so called microgravity environment of a satellite in orbit.
a) what is the max time interval for free fall if an experiment package were to fall the entire 146 m?

5.45 s [correct]

b) Actual NASA specs allow for a 5.24 s drop interval. How far do the packages in 5.24 s?

135m [correct]

c) What is their speed at 5.24 s?

51.5 m/s [correct]

d) What constant acceleration would be required to stop an experiment package in the distance remaining in the that shaft after its 5.24 fall?

121 m/s^2 [incorrect]

Homework Equations



Kinematic equations


The Attempt at a Solution



Presumably, since the shaft is 146 m tall and the package has already fallen 135 m there are 11 m left for the package to stop.

deltaX = 11 m : Vi = 51.5 m/s : Vf = 0 m/s

this information should be enough to calculate Ax for the second part of the problem but Webassign says my answer is wrong. Any ideas guys?

Problem 2

Homework Statement



An inquisitive physics student and mountain climber climbs a 48.0 m cliff that overhangs a calm pool of water. He throws two stones verticall downwar, 1.00 s apart and observes that they cause a single splash. The first stone has an initial speed of 1.92 m/s downward. Assume the downward direction is positive.

(a) How long after release of the first stone do the two stones hit the water?

2.93 s [correct]

(b) What initial velocity must the second stone have if they are to hit simultaneously?

16.1 m/s [incorrect]

(c) What is the speed of each stone at the instant the two hit the water?

first stone

30.7 m/s [correct]

second stone

34.62 m/s [correct]


Homework Equations



Kinematic equations


The Attempt at a Solution



This seems like a pretty straightforward problem but I can't seem to get question B right even though I used the value for question B to get the correct answer for question D.
 
Physics news on Phys.org
b)The package is accelerating down wards with g. Your answer is true if the package is moving with uniform velocity. So net acceleration should be ...?
 
131 m/s^2 in the upward direction to compensate for g, which is incorrect according to Webassign
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top