Analysis of Rocket Launch at 53 Degrees

In summary: What would the vertical velocity be at the top of the arc? (That is, when the vertical displacement is at its maximum). How about the horizontal velocity? What is the vertical velocity at the end of the flight? How about the horizontal velocity at the end of the flight?In summary, the problem involves finding the maximum altitude reached by a rocket launched at 53 degrees with an initial speed of 100 m/s and an acceleration of 30.0 meters per second squared for 3.00 seconds along its initial line of motion before its engines fail and it moves as a projectile. The equations of linear motion must be used to find
  • #1
DeagleBeagle
6
0

Homework Statement


A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. The rocket moves for 3.00s along its initial line of motion with an acceleration of 30.0 meters per second squared. At this time, its egines fail and the rocket proceeds to move as a projectile.



Homework Equations


1. Find the maximum altitude reached by the rocket
2. Find the total time of flight
3. Find the horizontal range



The Attempt at a Solution


Well I wrote down the problem and the main thing I need to figure out is a formula for related problems which give an angle and time something goes into the air with acceleration and ask for how far does it go up into the air after the power stops. I calculated for the first 3 secs it went 480 feet in the air (130+160+190). And it keeps going after that and eventually stops so I figure I need to use a formula with gravity. But I really want to figure this out because I missed a similar test question like this.

And I can't find out out the 2nd part of the problem I believe unless I figure out the first one. And for horizontal range, how does the angle affect the distance? Btw does anyone know how to make the degrees symbol or the little 2 above a number to symbolize it being squared?

 
Physics news on Phys.org
  • #2
First things first: Welcome to PhysicsForums, DeagleBeagle!

How did you get 480 feet? Did you take into account the flight angle, 53 degrees? What formula did you use? (Show your work, please. We have a template on purpose. You need that information to arrive at the right answer. We need that information so we can see where you went wrong.)

You are right in that you need the first part to get the answer to the second part. Breaking a problem down into more manageable sized pieces is often a good strategy.
 
  • #3
No, sorry I forgot about the angle affecting it. The feet came from the acceleration of 30 meters per second. 100 meters squared was the initial velocity and so after one second I figured 130, then 160, then 190. And the 3 seconds are up now, and I added those all together and got 480, but I know it is higher if you exclude the angle and assume it is 90, but it is 53, so I need to do it right. I didn't use any particular formula, just have no clue what to do, so I did that in my head.
 
  • #4
That's not how acceleration works. What's the right formula?
 
  • #5
I dunno, that is what I am trying to figure out.
 
  • #6
If this is not homework, it sure looks like a homework problem. If it is a homework problem, you probably have a book. What does the book say?
 
  • #7
It is very late. If DeagleBeagle comes back, someone else feel free to take over here.
 
  • #8
DeagleBeagle said:
I dunno, that is what I am trying to figure out.

What school are you attending? I'd like to get a list of your learning materials from you and e-mail your professor about how you do not seem to have adequate learning materials. Er, or do you have them just fine, and have not cracked the textbook open yet? You are not the first student to have been lazy and have asked for help here before. Do some work...

http://en.wikipedia.org/wiki/Equations_of_motion


.
 
  • #9
Hey hey, I didn't come here to be lazy. I came here to learn. I read all of the chapter relating to the problem and other ones I am doing but they don't cover specific problems perfectly and I have a hard time understanding it. I am in high school, not college. And the book just told me what the problem was, I don't know what you guys mean.
 
  • #10
DeagleBeagle said:
Hey hey, I didn't come here to be lazy. I came here to learn. I read all of the chapter relating to the problem and other ones I am doing but they don't cover specific problems perfectly and I have a hard time understanding it. I am in high school, not college. And the book just told me what the problem was, I don't know what you guys mean.

Fair enough. We will try to help, but you need to work with us. Did you reak the wikipedia article that I linked to in my last post? Those are the kinematic equations of motion for constant accelerations, which applies to your problem.

What is the equation that relates velocity and acceleration? Can you say how you think it might be applied in this problem?
 
  • #11
I read the whole article and I think the relevant formulas are in the equations of circular motion. I know I need to find the W, but I believe I have to take into account gravity and the angle so I don't know if any of these formulas apply. I think it may use this formula but I don't understand how they ask for time, a friend glanced at the problem and said you could use this formula: Dy = Vo Sin pheta t + 1/2gt^2
 
  • #12
DeagleBeagle said:
I read the whole article and I think the relevant formulas are in the equations of circular motion. I know I need to find the W, but I believe I have to take into account gravity and the angle so I don't know if any of these formulas apply. I think it may use this formula but I don't understand how they ask for time, a friend glanced at the problem and said you could use this formula: Dy = Vo Sin pheta t + 1/2gt^2

No, use the equations of linear motion, not circular motion.

Try doing some simpler versions of the problem first. Just do it with an initial velocity and no engine acceleration. The rocket (or simplify it to a particle) will follow a parabolic arc, with its horizontal velocity constant, and its vertical velocity slowing and turning around negative (due to the downward acceleration of gravity). Write the equations for that motion, to be sure that you understand how to work with them.

Then add in the acceleration time of the actual question. That will affect the rocket's velocity during the acceleration time. It will affect it in both the x and y directions, since the rocket is aimed at an angle.
 
  • #13
Oh, face-palm*, the formula's I gave were for a different problem, was looking at the wrong one. But ok sorry about that and the teacher printed out answer sheets for everyone to study how to do the homework. Btw the homework is not mandatory or graded, just a way to get practice.

Well for the first 3 seconds she (my teacher) used deltaX = VoT+1/2at2 to find I guess how high the rocket was when the engine failed. And she did: deltaX = 300 + 135 = 435m. And then she took the 435 number and did the cos of 53 degrees times that and got 262m for deltaX. And then for deltaY she took 435sin53 and got 347m.

I guess she did the first part to find how far it would go if just straight up? or what? And then seemed to make the number smaller by multiplying it by the cos or sin and the angle. This must be the real height because she ignored the 435 completely in the following work.

And then she used this formula: Vf = Vo+at and plugged in: Vf=100+90, Vf=190m/s . And then used the 190 times the cos or sin and the angle to find out the meters per second of Vx and Vy, and she got Vx=114m/s, and Vy=152m/s. Then to find time she used this formula: Vf=Vo+at, and then wrote: 0=152+(-9.81)(t), -152/-9.81= t = 15.5 secs .

And since we had to find the height she used (Vf+Vo/2)t=deltaY, then (152+0/2)(15.5) = 1178m. Ok and for the first part of the problem, A), she added the 1178m + 347m=1525m.
For the 2nd part, she wrote: Free Fall from 1525=1/2gt^2=17.6sec and the final answer she got with this: total jump=17.6+15.5= 36.1sec
3rd part: 17.6+15.5=33.1sec, DeltaX=Vt=(114)(33.1)=3773.4m, DeltaXt=3773.4+262m= 4035.4m. But why was the seconds 33.1? Why was the 3 seconds taken off? Is it because the 262 that was the distance it apparently went in 3 secs was already added on in the later portion of the problem?
 

What is the purpose of analyzing rocket launches at 53 degrees?

The purpose of analyzing rocket launches at 53 degrees is to understand the effects of launching a rocket at a specific angle on its trajectory, flight path, and overall performance. This information is important for optimizing rocket design and ensuring successful launches.

How is the analysis of rocket launches at 53 degrees conducted?

The analysis of rocket launches at 53 degrees is typically conducted using mathematical models and simulations. These models take into account factors such as atmospheric conditions, rocket design, and launch angle to predict the trajectory and performance of the rocket.

What are the potential benefits of launching a rocket at 53 degrees?

Launching a rocket at 53 degrees can have several benefits, such as achieving a specific orbital trajectory, minimizing the impact of atmospheric conditions on the rocket's flight path, and reducing fuel consumption. It may also be necessary for launching from certain geographical locations.

Are there any challenges or risks associated with launching a rocket at 53 degrees?

Yes, there are potential challenges and risks associated with launching a rocket at 53 degrees. These may include increased stress on the rocket's components due to the angle of launch, potential interference with other satellites or objects in orbit, and limitations on launch locations.

How does the analysis of rocket launches at 53 degrees contribute to the field of rocket science?

The analysis of rocket launches at 53 degrees is crucial for advancing the field of rocket science. It provides valuable data and insights that can be used to improve rocket design, optimize launch strategies, and ensure the safety and success of future launches. This information is also important for space exploration and other applications of rocket technology.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
380
  • Introductory Physics Homework Help
Replies
1
Views
883
  • Introductory Physics Homework Help
2
Replies
53
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
2
Replies
36
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
957
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
20K
Back
Top