## Hard Projectile Motion Problem.

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?

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 Recognitions: Homework Help 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.
 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.

Recognitions:
Homework Help

## Hard Projectile Motion Problem.

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 .

 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.

Recognitions:
Homework Help
 Quote by 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.
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.

 Recognitions: Homework Help 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.

Recognitions:
Homework Help
 Quote by radou 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.

Recognitions:
Homework Help
 Quote by Päällikkö 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.

 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?
 Recognitions: Homework Help 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?
 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??
 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??
 Recognitions: Homework Help 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?
 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!!
 In the equation V=Vo+at Vo would be the original/initial velocity right? So Vo=100m/s??

Recognitions:
Homework Help
 Quote by AznBoi 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.