Hard Projectile Motion Problem.

[AznBoi]

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!

 

[Päällikkö]

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.

 

[AznBoi]

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.

 

[Päällikkö]

Let's examine the problem while t < 3s:
The long way:
x₀ = 0
y₀ = 0
v_x,0 = 100 cos(53)
v_y,0 = 100 sin(53)
a_x = ?
a_y = ?

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

Actually, you don't really need to solve for v_x(t) or v_y(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 .

 

[AznBoi]

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.

 

[radou]

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.

 

[Päällikkö]

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.

 

[Päällikkö]

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.

 

[radou]

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:

 

[AznBoi]

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?

 

[Päällikkö]

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?

 

[AznBoi]

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??

 

[AznBoi]

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??

 

[Päällikkö]

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?

 

[AznBoi]

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!

 

[AznBoi]

In the equation V=Vo+at

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

 

[radou]

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.

 

[Päällikkö]

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 .

 

[Päällikkö]

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.

 

[AznBoi]

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?

 

[Päällikkö]

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.

 

[AznBoi]

Wow, I really appreciate you homework helpers helping me out in a long problem like this. Thanks to all of you! For this problem, especially Paallikko and Radou. I'll solve the rest on my own now.

 

[AznBoi]

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?

 

[Päällikkö]

I think I mentioned about that before. Anyways, you are supposed to use Ay there, can you figure out why?

 

[AznBoi]

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.

 

[Päällikkö]

Gravity does have an effect. It is, however, already included in the given acceleration.

Offtopic: Don't let this confuse you: If the problem was stated differently, eg. the mass of the rocket (shot in some angle theta) was some given m, and its engines could exert some force F, you'd have to take the gravity into account, too.

 

[AznBoi]

Oooh, I see. Because the rocket is accelerating, yeah the gravity has already been included. I understand what you mean. Thanks so much!

 

[AznBoi]

Ok, I've gotten the answers for t<3, x= 261.765m and for y=347.4m

Are they correct? I have the solution for the answers but for the a) maximum altitude reached by the rocket it is 1.52 x 10^3 m. Isn't that 1520m?? For this answer it is only 350 meters, is it possible for it to reach 1170m during the parabloic motion afterwards?

Also for the total displacement the answer is 4.05 x 10^3 m =/

 

[Päällikkö]

You could do the maximum height (after t = 3s) easily with conservation of energy. The answer 347.4 m fits the max altitude of 1520 m perfectly.

I didn't check the range, but it looks reasonable.

 

[AznBoi]

Whats the conversion of energy? I only know how to do it the long way lol. I'm a precal student so I don't know any calculus yet, I wish I did though. =P I'm just going to do the same thing again? lol

 

[Päällikkö]

Conservation of energy would be:
$$\frac{1}{2}mv^2 = mgh$$

if you're unfamiliar with that, just do it the long way . It's good practice anyways.

 

[AznBoi]

yeah I haven't learned that equation yet. Thanks for all your help again!