# Could use some help on physics (mechanics problem help)

1. May 4, 2004

### shivas

i was wondering if someone on this board would kindly bestow their knowledge onto this problem... thank you..

A model rocket launches with initial net upward acceleration of 1.5g where g is the absolute value of the acceleration due to gravity. This accerlation is maintained for 3 secs. Adter the 3 secs has passed the rocket coasts and eventually falls back to earth. At what time does the rocket reach its max height??

-- thanks!

2. May 4, 2004

### Integral

Staff Emeritus
What have you done?
What equations have you been given that might be of use? There are several different ways of solving this problem depending on the level of your Math, show us some of what you know about it. That will help us know where to start.

3. May 4, 2004

### shivas

well im thinkin i should use kinematics equations to solve this.... but not sure which one

i know how to do calculus...

4. May 4, 2004

### Parth Dave

kinematics equations are all you need.

First of all, what can you say about the velocity when the rocket reaches its max height? see if that helps.

5. May 4, 2004

### shivas

velocity is zero at maX, right?

6. May 4, 2004

### Parth Dave

correct. Now what formula do you have to find velocity?
and what variables do you know in the formula?

7. May 5, 2004

### ShawnD

Break it into 2 parts, accelerating and decelerating. First, we know the time for the acceleration part, now find the velocity for it.

$$V = at$$

$$V = (1.5g)(3)$$

$$V = 4.5g$$

Now for the second part. We know the final velocity is 0 but we don't know the time, so lets find the time.

$$V_f = V_i + at$$

Vf is 0, Vi is the V from part one, a is -g (since it's the opposite direction of the velocity) and t is just t.

$$0 = 4.5g + (-g)(t)$$

$$gt = 4.5g$$

$$t = 4.5$$

So you know the time from part one is 3s and the time from part 2 is 4.5 seconds. Now add em together

$$t = 3 + 4.5$$

$$t = 7.5$$ seconds

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