Set Upward Accel. Then FreeFall ?

[moobers]

Homework Statement

For some reason, I just can't seem to understand this question. I didn't think it would be this bad. Anyways, A Rocket fires from rest with an upward acceleartion of 30 m/s^s for 2 seconds. After this time the engine shuts off and the rocker freely falls back to the surface of the Earth. ( use g = 10 m/2^s)
a) Draw a v vs t graph. I've done that already. ;'D
b) Determine the height of the rocket when the engines shut off. Done.
c) Determine the Maximum height that the rocket reaches.<--- Okay. THis is the one. <--- For some reason I'm thinking that at the crest of the curve/line/ place before it falls its V = 0, but I have no idea how to figure out how much farther/higher it goes. And that is my problem.



2. The attempt at a solution

I've tried using, x = 1/2at^2+VoT+Xo and V^2 = Vo2+2AX.
But neither are right. Or I'm doing something wrong. D:
THanks in advance for any help or comments. :'D

Homework Statement

 

[neutrino]

You're right in saying that the at the crest of its trajectory the rocket's velocity is zero.

The rocket's upward motion can be split in two parts, 1. The part with the thrust and 2. The part where its still moving up with only the influence of gravity. For the second part, the acceleration is known (what are the forces acting on it?), and so is the initial velocity (look at the graph) and final velocity. Now use "V^2 = Vo2+2AX" to find the distance traveled during the second part of it's upward journey.

 

[m2003]

For some reason, I just can't seem to understand this question. I didn't think it would be this bad. Anyways, A Rocket fires from rest with an upward acceleration of 30 m/s^2 for 2 seconds. After this time the engine shuts off and the rocket freely falls back to the surface of the Earth. ( use g = 10 m/s^2)
a) Draw a v vs t graph. I've done that already. ;'D
b) Determine the height of the rocket when the engines shut off. Done.
c) Determine the Maximum height that the rocket reaches.

As a scientist, it is important to approach problems with a clear and logical mindset. In this case, it seems like you have already made some progress by drawing the v vs t graph and determining the height of the rocket when the engines shut off. However, it is important to understand that the maximum height of the rocket can be calculated using the equations of motion and the principles of free fall.

To begin, let's recall the equations of motion for constant acceleration:

v = u + at
s = ut + 1/2at^2
v^2 = u^2 + 2as

In this case, the rocket has an initial velocity of 0 m/s (since it starts from rest) and an upward acceleration of 30 m/s^2. Therefore, the equations become:

v = 0 + 30(2)
s = 0(2) + 1/2(30)(2)^2
v^2 = 0^2 + 2(30)s

Solving for the variables, we get:

v = 60 m/s
s = 60 m
v^2 = 2(30)(60) = 3600 m^2/s^2

Now, we can use the principle of free fall to determine the maximum height of the rocket. This principle states that the maximum height reached by an object in free fall is equal to the initial height plus the change in height during free fall. In this case, the initial height is 60 m and the change in height is 60 m (since the rocket goes up and then falls back to the ground). Therefore, the maximum height reached by the rocket is 60 m + 60 m = 120 m.

I hope this explanation helps you to better understand the problem and how to approach it using scientific principles