Kinematics Involving Ball, What did I do wrong?

[dantechiesa]

Homework Statement

Your cousin has built a model rocket that she wishes to launch in your backyard. You're a little worried by this, and ask her for details.
She tells you, when I launch the rocket straight upwards, the acceleration will be 3.5g = (3.5*9.8m/s^2) for 4.3 seconds, then burn out.
A) What is the max height the rocket will reach
and specifically, what I need help with, what will the time from launch until the rocket returns to the ground.

Homework Equations

V^2 = V0^2 +2ad
d = v0t + .5at^2

3. The Attempt at a Solution
So, what I am having trouble with is calculating the time the rocket spends in the air after its acceleration fuel burns out (after 4.3 seconds)

Phase 1: the initial 4.3 seconds
Phase 2: the remaining time it spends in the air
Phase 3: the rocket falls to the ground.

Right when the first phase ends, I was able to calculate its
Vf = 147.49m/s^2 (subsequently is the V0 for phase 2)
d = 317.1035m and phase 2 distance = 1102.5m

However, when I try to calculate time, the quadratic formula isn't giving the correct time.
I input into my calculator
1102.5 = 147.49t - 4.9t^2
0 = -4.9t^2 + 147.49 - 1102.5
This is giving 2 positive roots.
What have I done wrong?

 

[Doc Al]

phase 2 distance = 1102.5m

I assume this is the additional distance the object rises.

To solve for the total time after the rocket burns out, consider the initial and final positions for that segment of the motion.

 

[Lori]

Homework Statement

Your cousin has built a model rocket that she wishes to launch in your backyard. You're a little worried by this, and ask her for details.
She tells you, when I launch the rocket straight upwards, the acceleration will be 3.5g = (3.5*9.8m/s^2) for 4.3 seconds, then burn out.
A) What is the max height the rocket will reach
and specifically, what I need help with, what will the time from launch until the rocket returns to the ground.

Homework Equations

V^2 = V0^2 +2ad
d = v0t + .5at^2

3. The Attempt at a Solution
So, what I am having trouble with is calculating the time the rocket spends in the air after its acceleration fuel burns out (after 4.3 seconds)

Phase 1: the initial 4.3 seconds
Phase 2: the remaining time it spends in the air
Phase 3: the rocket falls to the ground.

Right when the first phase ends, I was able to calculate its
Vf = 147.49m/s^2 (subsequently is the V0 for phase 2)
d = 317.1035m and phase 2 distance = 1102.5m

However, when I try to calculate time, the quadratic formula isn't giving the correct time.
I input into my calculator
1102.5 = 147.49t - 4.9t^2
0 = -4.9t^2 + 147.49 - 1102.5
This is giving 2 positive roots.
What have I done wrong?

Hmmm. I'm a fellow student wanting to help out!

What i did was treat the second phase's initial y velocity as 0 because the rocket "burns out" which means its velocity is 0. With the distance being 317.1035 and initial velocity being 0 and the acceleration is positive 9.8 m/s^2, i calculated t=8.04 secs for the second phase. time for phase1 + phase 2 = 12.35 seconds.

I'm am not entirely sure if that's right though!

 

[Dr.D]

What i did was treat the second phase's initial y velocity as 0 because the rocket "burns out" which means its velocity is 0.

Not quite so. At burnout the rocket has the altitude and velocity attained to that point. So the post-burn motion starts with a nonzero position and a nonzero velocity.