Hard Projectile Motion Problem.

  • Thread starter AznBoi
  • Start date
  • #1
471
0
Ugh. My last projectile motion problem. I just don't get one thing in this problem. The initial velocity and the acceleration. Please help me with this problem:

A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100m/s. It moves for 3s along its initial like of motion with an acceleration of 30m/s^2. At this time its engines fail and the rocket proceeds to move as a projectile. Find a) the maximum altitude reached by the rocket, b) its total time of flight, and c) is horizontal range.

My work so far:
Alright, I have a basic sketch of the problem. However, I'm not quite sure on where to get started. Basically there is a linear line that extends for 3s and has a X displacement. Then the rocket drifts a parabolic motion until it reaches the ground.

In order to calculate the displacement of the 3s, you need to find the x and y velocity components right? Here is what confuses me, how do you find the components with both an intial velocity and acceleration?

In my last problem I only had the inital velocity, and finding the velocity components was fairly easy. However, now I have the initial velocity+ the acceleration. How am I suppose to find the components of that?

Do I add them up and use cosine, sine? Please help me get started. Thanks a lot! :smile:
 

Answers and Replies

  • #2
Päällikkö
Homework Helper
519
11
In pretty much every projectile motion problem you have there's a constant acceleration g in the y direction. You usually have 0 acceleration in the x direction. You then go and deal with the x and y equations of motion separately. The same applies in this problem: you can separate the problem into motion in the x direction and motion in the y direction.

You're given the total acceleration of the rocket, surely you can break it up to x and y directions using sines and cosines?

You should deal with the problem in 2 phases. First the motion of the rocket before t = 3s, and then the motion of the rocket after t = 3s.
 
  • #3
471
0
So do you add the initial velocity to the acceleration then find the x and y components of that?

Vy= 130m/s(Sin 53)
Vx= 130m/s(Cos 53)

I know I'm doing something wrong.:uhh:
 
  • #4
Päällikkö
Homework Helper
519
11
Let's examine the problem while t < 3s:
The long way:
x0 = 0
y0 = 0
vx,0 = 100 cos(53)
vy,0 = 100 sin(53)
ax = ?
ay = ?

So, just as you've probably done a million times before:
x(t) = ?
y(t) = ?
vx(t) = ?
vy(t) = ?

Actually, you don't really need to solve for vx(t) or vy(t) from the equations above (you easily get them by solving the total velocity v = at, and breaking that into x and y components), but it makes good practice :smile:.
 
Last edited:
  • #5
471
0
Hmm.. so the initial velocity and the acceleration have two different x and y componenets?? I still don't get where this is going. After I find the x and y components for the velocity 100m/s, what do I do? Does the acceleration have x and y components too?? This is the first problem I have encountered that has something like this. If it didn't have the accleration, it would be easier. Thats what I'm confused about. Thanks.
 
  • #6
radou
Homework Helper
3,120
7
AznBoi said:
Hmm.. so the initial velocity and the acceleration have two different x and y componenets?? I still don't get where this is going. After I find the x and y components for the velocity 100m/s, what do I do? Does the acceleration have x and y components too?? This is the first problem I have encountered that has something like this. If it didn't have the accleration, it would be easier. Thats what I'm confused about. Thanks.

The acceleration in the y-direction is gravitational, and in the x-direction it equals 30 m/s^2, until the time the rocket engine fails.
 
  • #7
Päällikkö
Homework Helper
519
11
Yes, the acceleration does have x and y components too. Acceleration, like velocity, is a vector. You've had acceleration in your projectile motion problems before. Although then it's been the acceleration caused by gravitation, and it has only been directed in the y direction.

The problem would "reduce" into a simple projectile motion problem if you were given the velocity and the displacement in the x and y directions at t = 3s, right? That's where we're headed: We need to figure out what those values are.
 
  • #8
Päällikkö
Homework Helper
519
11
radou said:
The acceleration in the y-direction is gravitational, and in the x-direction it equals 30 m/s^2, until the time the rocket engine fails.
Unless I've misunderstood something, you are wrong. The rocket has an acceleration of 30 m/s2, into the given direction of 53 degrees above the horizontal. That already accounts for the acceleration caused by gravity.
 
Last edited:
  • #9
radou
Homework Helper
3,120
7
Päällikkö said:
Unless I've misunderstood something, you are wrong. The rocket has an acceleration on 30 m/s2, into the given direction of 53 degrees above the horizontal. That already accounts for the acceleration caused by gravity.

You're right, I misunderstood it. Sorry. :shy:
 
  • #10
471
0
Ok, I've found the x and y components of the initial velocity 100m/s

Vx=60.18m/s
Vy=79.86m/s

So I can put 30m/s^2 in a parallelogram, as the resultant??

Then the x and y components of acceleration are:

Vx= 30m/s^2(cos 53)=18.05m/s^2
vy=30m/s^2(sin 53)=23.96m/s^2

Now what do I do with those components?? I can't add them together or anything right?
 
  • #11
Päällikkö
Homework Helper
519
11
You used a bit strange symbols to denote the acceleration in x and y directions (Vx and Vy, the same you used for velocities). A better choice would be Ax and Ay. Anyways, you're probably familiar with the equations:
x = x0 + v0t + (1/2)at2
v = v0 + at

Can you see how to use them to get the velocities and displacements at t = 3s, the instant the rocket's engine fails?
 
  • #12
471
0
Oh so when you find the distance, x, you would substitue the (X component) of acceleration for Ax??

So,

X=Voxt+.5(Ax)t^2

X=(60.18m/s)(t)+.5(18.05)t^2

So I just need to find t now right??
 
Last edited:
  • #13
471
0
Yea, you are right, I should use Ax and Ay for the accel components.

For the time, how do I find it with V=Vo+at??

Would V=100+ 30m/s^2(3s)? therefore would it be 190m/s?

Would Vo be 100m/s?

And the A would be 30m/s^2??
 
  • #14
Päällikkö
Homework Helper
519
11
Well t is obviously 3s, that's the time the problem turns into a normal projectile motion problem, remember? The equations we currently have do not hold after the engine fails, as there will be no acceleration in the x direction. This is why we are currently only considering t < 3s.

The equation you wrote above is correct. Do the same for the y direction, and for the velocity.

EDIT: All of the equations written above are correct. I'm, however, not quite sure what you mean by V0 = 100 m/s?
 
  • #15
471
0
Lol, yeah the time is 3s xP I forgot about that.

Oh, so if I do the same for Y=Yo+Vyot+.5(g)(t)^2 I will get the height of the 3 second period of time?? Therefore it would look like a triangle right?? The distance X on the bottom leg, and the height Y on the opposite leg?

Thanks a lot, I'm starting to get this. I was just confused with the acceleration. So after the engine runs out, the problem would have a constant velocity Vxo. Ok I will try to solve the rest of the problem from here on. Thanks again to you both!! :tongue:
 
  • #16
471
0
In the equation V=Vo+at

Vo would be the original/initial velocity right? So Vo=100m/s??
 
  • #17
radou
Homework Helper
3,120
7
AznBoi said:
In the equation V=Vo+at

Vo would be the original/initial velocity right? So Vo=100m/s??

Every equation you write has to be an equation for some direction, x or y. So, Vx = Vox + ax*t and Vy = Voy + ay*t.
 
  • #18
Päällikkö
Homework Helper
519
11
Indeed, it'll look like the triangle you described (while t < 3s, after that it'll take the paraboloidic form of "traditional" projectile motion).

You've made one mistake: acceleration in y direction while t < 3 s is not g.
After the engine runs out, the rocket will also have velocity in the y direction.

Happy problemsolving :smile:.
 
  • #19
Päällikkö
Homework Helper
519
11
radou said:
Every equation you write has to be an equation for some direction, x or y. So, Vx = Vox + ax*t and Vy = Voy + ay*t.
Well actually, this is a case you don't have to break into components as the acceleration is parallel to velocity. So AznBoi's equation is correct.
 
  • #20
471
0
OH!! I get it, oops lol. The equation V=Vo+at is to determine the final velocity, or the initial velocity of the parabolic motion after the engine stops right??


V(final)=Vo(Initial)+a(acceleration)t(time)

V=100m/s+(30m/s^2)(3s)

So therefore the final Velocity would be: 190m/s

That would be the velocity Right After the engine stops. The initial velocity of the second part of the problem, the regular parabola/projectile motion part. That velocity would remain the same until it hits the ground right?
 
  • #21
Päällikkö
Homework Helper
519
11
Yes, the final velocity of the first part of the problem (190m/s^2) is the initial velocity of the second part. Displacements behave the same. The only thing that changes is acceleration.
 
  • #22
471
0
Wow, I really appreciate you homework helpers helping me out in a long problem like this. Thanks to all of you!! :tongue: For this problem, especially Paallikko and Radou. I'll solve the rest on my own now. :smile:
 
  • #23
471
0
Wait, one more question. =P

When you find the height of t<3, you use the equation:
Y=Yo+Voyt+.5(g)(t)^2 right??

Ok, for (g) do you put (-9.8m/s^2) or Ay??
If you don't put Ay in g, where do you substitute it?
 
  • #24
Päällikkö
Homework Helper
519
11
I think I mentioned about that before. Anyways, you are supposed to use Ay there, can you figure out why?
 
  • #25
471
0
Oh, yeah I was wondering why g(-9.8m/s^2) didn't make sense in that equation. It is because the acceleration takes up for the force of gravity. lol, I don't really know how to say it correctly but the objects is accerelating upwards and it is not the y motion, so the gravity force doesn't act upon it? If there was no acceleration, the force of gravity would act upon it and it would be (-9.8m/s^2). Right? Thanks again. :smile:
 

Related Threads on Hard Projectile Motion Problem.

  • Last Post
Replies
9
Views
11K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
18
Views
3K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
544
  • Last Post
Replies
11
Views
875
Top