# Rocket Problem

1. Sep 14, 2009

### DeagleBeagle

1. The problem statement, all variables and given/known data
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.

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

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 14, 2009

### D H

Staff Emeritus
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. Sep 14, 2009

### DeagleBeagle

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. Sep 15, 2009

### D H

Staff Emeritus
That's not how acceleration works. What's the right formula?

5. Sep 15, 2009

### DeagleBeagle

I dunno, that is what I am trying to figure out.

6. Sep 15, 2009

### D H

Staff Emeritus
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. Sep 15, 2009

### D H

Staff Emeritus
It is very late. If DeagleBeagle comes back, someone else feel free to take over here.

8. Sep 15, 2009

### Staff: Mentor

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. Sep 15, 2009

### DeagleBeagle

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. Sep 15, 2009

### Staff: Mentor

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. Sep 15, 2009

### DeagleBeagle

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. Sep 15, 2009

### Staff: Mentor

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. Sep 16, 2009

### DeagleBeagle

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?