# Homework Help: Finding Time From Acceleration

1. Feb 9, 2009

### lazyboi605

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

A model rocket takes off from ground level accelerating upward at 4.0 g. This upward acceleration lasts for 33 s. Afterward the rocket continues upward, eventually stops rising, then falls back to the ground.

How much time passes from the initial upward acceleration stoping to the rocket returning to the ground?

2. Relevant equations

3. The attempt at a solution

2. Feb 9, 2009

### LowlyPion

What are your thoughts on how to solve it?

3. Feb 9, 2009

### lazyboi605

I am using velocity/gravity, but it don't seem to work out.

4. Feb 9, 2009

### LowlyPion

That only gives you the time to max height.

If you want, you can figure that height and then get the time to fall in gravity from x = 1/2*g*t2

Then just add the 2 times together.

5. Feb 9, 2009

### lazyboi605

Would i need to minus 33, because i am trying to find the time passed from initial upward acceleration stopping.

6. Feb 9, 2009

### lazyboi605

Also i don't understand how to find the time to max height. because it will still move after it accelerate.

7. Feb 10, 2009

### LowlyPion

V = a*t
So when the engine shuts off, you are going at (4*g)*33 m/s
and dividing by g yields the additional time to max height = 4*33 s.

Now solve for X of engine shutoff X = 1/2 *4*g *t2
That's the point when the engine stops.

Figure the additional height from that point by
V2 = 2*g*x

Add the 2 heights - up to engine shutoff and from shutoff to 0 velocity at the top.

Using that height now as your X you have only to solve for the time using

X = 1/2*g*t2

That time to fall + time to max height from engine shut off is what they want.