Finding Max Altitude Reached by Rocket with Constant Acceleration

[CubeDude]

Homework Statement

A Rocket is initially at rest on the ground. The rocket flies off in a straight line at an angle 53.1 degrees above the horizontal with a constant acceleration of magnitude [tex]\textbf{g}[/tex]. The engines stop at a time [tex]\textbf{T}[/tex] after launch, after which the rocket is in projectile motion.

Ignoring air resistance, and assuming [tex]\textbf{g}[/tex] is indepedent of altitude; Find the maximum altitude reached by the rocket, your answer should be in terms of [tex]\textbf{g}[/tex] and [tex]\textbf{T}[/tex].

Homework Equations

y = y₀ + V_0yt + [tex]1/2[/tex]at[tex]^{2}[/tex]
V₂² = V₁² + 2aY
V₁ = V₀ + at

The Attempt at a Solution

]
I attempt to tackle this problem by seperately finding an expression for height after time [tex]\textbf{T}[/tex], and an expression for the maximum height of a projectile released with the velocity the rocket has at time [tex]\textbf{T}[/tex]. I would then take the maximum height to be the sum of these two expressions.

The question states that the rocket travels in a straight line. Am I right to assume that this means before time [tex]\textbf{T}[/tex] the vertical component of the rocket's acceleration is equal to the acceleration due to gravity but in the oposite direction?

If I assume the acceleration before time [tex]\textbf{T}[/tex] is zero, then the height reached by the rocket at time [tex]\textbf{T}[/tex] is given by:

y₁ = (V₀Sin[tex]\alpha[/tex])[tex]\textbf{T}[/tex]

If the rocket is not accelerating then at time [tex]\textbf{T}[/tex] the velocity will be equal to the rocket's intitial velocity.
Using:
V₂² = V₁² + 2aY₂
Taking V₂² = 0, and rearranging.
Y₂ = (V₀²Sin²[tex]\alpha[/tex]) [tex]1/2g[/tex]

The correct answer is independent of the Rocket's initial velocity. I don't see how I can cancel these out, so I must be going wrong somewhere.

I don't start my Physics course until October, but I'm trying to get a bit of a head start and this is the first problem I've become stuck on. Any hints to nudge me in the right direction would be greatly appreciated.

(I can't seem to make the alpha next to the Sin stay at the same level as the Sin. It should mean Sin of the angle alpha.)

 

[G01]

Hmmm. The wording of the question is a little strange. I don't think they mean that the rocket travels horizontally for the first T seconds, but rather that the rocket travels in a straight line at 53.1 degrees. In this case, I also think they mean that for those first T seconds, the rocket's acceleration is g in that direction (i.e. the direct of that line).

Does this make sense?

So, for the first part your going to have some upward acceleration equal to the y component of the acceleration in the 53.1 degree direction.

 

[CubeDude]

Thank you for the reply. I did originally assume that I would have to take the y component of the acceleration which would be gSinAlpha. Unfortunately on my first attempt I confused matters by still accounting for the downward acceleration due to gravity during the period before T. I also quite absentmindedly (a lesson I intend to bear in mind) ignored the fact that the initial velocity was zero.

Taking V₀ to be zero, and that the y component of the acceleration is gSinAlpha:

Y₁ + Y₂ = [tex]\textbf{g}[/tex][tex]\textbf{T}[/tex]² (Sin²Alpha + SinAlpha) [tex]\frac{1}{2}[/tex]

Substituting 53.1 degrees for Alpha provides the correct answer.

Thank you very much for your help.

 

[G01]

Thank you for the reply. I did originally assume that I would have to take the y component of the acceleration which would be gSinAlpha. Unfortunately on my first attempt I confused matters by still accounting for the downward acceleration due to gravity during the period before T. I also quite absentmindedly (a lesson I intend to bear in mind) ignored the fact that the initial velocity was zero.

Taking V₀ to be zero, and that the y component of the acceleration is gSinAlpha:

Y₁ + Y₂ = [tex]\textbf{g}[/tex][tex]\textbf{T}[/tex]² (Sin²Alpha + SinAlpha) [tex]\frac{1}{2}[/tex]

Substituting 53.1 degrees for Alpha provides the correct answer.

Thank you very much for your help.

Anytime.

Good job!