Try This
Elizabeth,
You have to break this problem up into parts. In the first time interval, the rocket is being acted on by the engines acceleration. Thats Interval 1. Then the engines fail and the rockets descent begins to slow, more and more until it is falling towards the earth. This is the second part (the freefall). First, find the time interval 1, which is how long it will take the rocket to accelerate itself to a altitude of [itex]950m[/itex].
This is given by:
[tex]\Delta y = V_0t + \frac{1}{2}at^2[/tex]
Now solve for [itex]t[/itex]. Tuck this answer away so that we can finish this part of the problem later.
Part (b) asks you to find the maximum height attained by the rocket. Remeber, the rocket propels itself to an altitude of 950m, and after that is when the engines fail and the rocket begins to slow down. We need to know how fast the rocket is going at the time of the engine failure. [itex]V_f[/itex]. Use this equation:
[tex]V_f = V_0 + at[/tex]
We know the initial speed of the rocket (from the catapult), we know its acceleration from the engines, and we know the time from our first answer in part (a).
Okay, so far so good. Now this equation comes to mind:
[tex]V^2_f = V^2_0 - 2g \Delta y[/tex]
Now let's plug in what we know so that we can get [itex]\Delta y[/itex], which is the height the rocket attains. (Note, this is the height from the 950m mark, not from the ground) Solve for [itex]\Delta y[/itex] and add the original 950 meters. This gives you the height from above the ground.
The last part of the problem is to find the velocity just before the rocket hits the earth. Let's treat the 950 meters point as the origin or point of reference. The Earth is 950 m BELOW this so we need to solve for when [itex]y = -950[/itex]. Use:
[tex]\Delta y = V_0t - \frac{1}{2}gt^2[/tex]
This will allow you to solve for [itex]t[/itex], which is the time for the second interval we talked about earlier. Add these times together and that should be part (a). Now, Last but not least, we need to find the final velocity.
[tex]V_f = V_0-gt[/tex]
Solve for [tex]V_f[/tex] where [tex]V_0[/tex] is the initial velocity the rocket reached at 950 meters, and [itex]t[/itex] is the time we just calculated. Remember, since the rocket will be falling down, your answer should be negative as velocity is a vector quantity (Unless you are using a different frame of reference than I was for this problem).
Hope this helps, I was working the problem quickly here at work. I will double check everything later.
On a side note, If you learn these concepts and learn how to *think* about the problem, the second problem you listed will come to you with greater ease than this one. SO GO STUDY!
