Early test flights for the space shuttle used a glider

  • Thread starter Thread starter kritzy
  • Start date Start date
  • Tags Tags
    Space Test
Click For Summary

Homework Help Overview

The discussion revolves around the physics of early test flights for the space shuttle using a glider, specifically focusing on calculating the landing speed in the absence of air resistance and determining the average force of air resistance experienced during descent.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between kinetic and potential energy to address the landing speed question, with some questioning the units used for acceleration due to gravity.
  • There is an attempt to calculate the landing speed by considering horizontal and vertical components of velocity, while others express uncertainty about incorporating air resistance and the effect of glide angle in their calculations.

Discussion Status

Some participants have made progress on the first part of the problem, while others are still grappling with the second question regarding air resistance. Multiple interpretations of the problem setup are being explored, particularly concerning the role of forces and angles in the calculations.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can use or the methods they can apply. There is also a focus on ensuring that calculations align with the physical principles involved.

kritzy
Messages
12
Reaction score
0

Homework Statement


Early test flights for the space shuttle used a "glider" (mass of 940 kg including pilot). After a horizontal launch at 470 km/h at a height of 4000 m, the glider eventually landed at a speed of 210 km/h . What would its landing speed have been in the absence of air resistance(in km/h)? What was the average force of air resistance exerted on it if it came in at a constant glide angle of 15 degrees to the Earth's surface?

Homework Equations


Ki + Ui = Kf + Uf
.5mv^{2}+mgh=.5mv^{2}

The Attempt at a Solution


I'm not so sure about the second question but here's what I did for the first one.
I changed the units so that the answer would be in km/h.
g=35.28 km/h
h=4 km
v^{2}=[2(.5mv^{2}+mgh)]/m
v^{2}=[2(.5(940)(470)^{2}+(940)(35.28)(4)]=221182.24
v=470km/h
Unfortunately, this answer is incorrect. Any help would be greatly appreciated.
 
Physics news on Phys.org


g is acceleration? The units could be kilometers per hour squared, rather than kilometers per hour.
 


In the absence of air resistance, the horizontal velocity will remain as it is.
Find its velocity when if falls through 4 km. The resultant of this velocity and the horizontal velocity will give the landing velocity.
 


Thanks. I got the first part. As for the second question, I'm not quite sure how to go about it. I think I'm suppose to use F=ma. So I caculated the acceleration.
a=(v^{2}-v^{2}_{0})/(2x)
a=(210^{2}-470^{2})/((2)(4))=-22100

F=ma=(940)(-22100)=-2077400

But that only gets me the force of the gilder and doesn't include friction. I was also thinking that I should get the sum of forces.So
F_{gilder}-F_{friction}=ma
I don't think that's quite right either. There suppose to be a 15 degree angle but I don't know where it goes. Can somebody help me?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 10 ·
Replies
10
Views
9K
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
9K