# Max altitude of a rocket

1. Sep 12, 2011

### l888l888l888

1. The problem statement, all variables and given/known data

A rocket is red vertically with an acceleration of 250 m/s2. After 30 seconds, the
rocket's motor shuts o . Find the maximum altitude achieved by the rocket and the
total time from take-o to return to the surface of the earth, assuming that the rocket's
design makes air resistance negligible.

2. Relevant equations

3. The attempt at a solution
d^2x/dt=a=acceleration.
dx/dt=at+c = velocity=v(t)
v(0)=0 implies c=0 so v(t)=at
x(t)=1/2*a*t*t +d. x(0)=0 implies d=0 so x(t)=x(t)=1/2*a*t*t
x(t)=(1/2)*a*t*t=1/2 * (250)*30*30=112500 meters
112500=1/2 * 9.8 * t$*t$. where t$is the time it takes to return to earth. t$= 151.52 secs
total time is t +t\$=181.52 secs.

Is this right? My classmates did it differently but I do not understand how I am wrong. please help!!!!!

2. Sep 12, 2011

### Delphi51

You are correct!

3. Sep 12, 2011

### l888l888l888

I hope so. My classmates are saying that after 30 seconds when the motor shuts off the rocket still continues upward for a while under the force of gravity (ie the rocket does not immediately return to the ground). if that is the case how would i do this problem differently?

4. Sep 12, 2011

### sjb-2812

Are you sure? After the thrust has finished, the rocket will be travelling quite fast, and will not instantaneously start moving downwards.

OP, what is the speed after 30 seconds of acceleration? Now, you have an initial velocity, an acceleration, a displacement, so you can calculate a time.

5. Sep 12, 2011

### l888l888l888

that is what my classmates are saying. but i dont know exactly how to change wat i did. can you help me?

6. Sep 12, 2011

### Delphi51

Caught me!
We have to calculate the velocity at time 30 s, how long it continues to go up before reaching speed zero, and how far it goes up in that time.

7. Sep 13, 2011

### sjb-2812

You can split this into two parts
a) whilst the trust is going on
b) after it is turned off.

a) initial velocity = 0, initial acceleration = 250 m/s2, time = 30 s.

Displacement = x Final velocity = y

b) initial velocity = y, acceleration = -g, displacement = -x

Time = w

b)' When the height is at a max, the velocity must be 0
initial velocity = y, acceleration = -g, final velocity = 0

Displacement = v