Kinematics how high does it go problem. Check

[Eric_meyers]

Homework Statement

A toy rocket, launched from the ground, rises vertically with an acceleration of 33 m /s^2 for 13 s until its motor stops
The acceleration of gravity = 9.8 m/s^2
Disregarding any air resistance, what maximum height above the ground will the rocket achieve? Answer in units of km.

G = -9.8 m/s^2
A = 33 m/s^2 with T = 13 seconds

Unknown
x - xi = ?

Homework Equations

x - xi = vi * t + 1/2 A T^2

V = A * T

V = Vi - gt

The Attempt at a Solution

Ok, so first I broke this problem up into two pieces, I checked the distance it traveled while accelerating by using the first equation I provided... the Vi * t = 0 since my Vi was 0 and the rest I got --- 1960.4 m

then I saw what velocity it was going at that time and I got 301.6 m/s

So then I needed to find the distance it traveled from 301.6 m/s to 0 m/s so I used the third equation and set V = 0 to find T which equals 30.7755102 seconds. Then I reused the 1st equation to find the distance of this part

By plugging g in for my acceleration component of course, I got 4,640.946939 m + my first part 6,601.346939 m / 1000 = 6.601.34 km


However this answer came up as wrong, ( I still have 6 tries left!) I can't see where I went wrong? I believe it's in the 2nd part finding how much distance the velocity goes until it reaches 0. Thanks for your help!

 

[LowlyPion]

Homework Statement

A toy rocket, launched from the ground, rises vertically with an acceleration of 33 m /s^2 for 13 s until its motor stops
The acceleration of gravity = 9.8 m/s^2
Disregarding any air resistance, what maximum height above the ground will the rocket achieve? Answer in units of km.

G = -9.8 m/s^2
A = 33 m/s^2 with T = 13 seconds

Unknown
x - xi = ?

Homework Equations

x - xi = vi * t + 1/2 A T^2

V = A * T
V = Vi - gt

The Attempt at a Solution

Ok, so first I broke this problem up into two pieces, I checked the distance it traveled while accelerating by using the first equation I provided... the Vi * t = 0 since my Vi was 0 and the rest I got --- 1960.4 m

then I saw what velocity it was going at that time and I got 301.6 m/s

So then I needed to find the distance it traveled from 301.6 m/s to 0 m/s so I used the third equation and set V = 0 to find T which equals 30.7755102 seconds. Then I reused the 1st equation to find the distance of this part

By plugging g in for my acceleration component of course, I got 4,640.946939 m + my first part 6,601.346939 m / 1000 = 6.601.34 km


However this answer came up as wrong, ( I still have 6 tries left!) I can't see where I went wrong? I believe it's in the 2nd part finding how much distance the velocity goes until it reaches 0. Thanks for your help!

How fast was the rocket going when the engines stopped?

 

[Eric_meyers]

I calculated it to be 301.6 m/s

This is the only problem I'm currently missing but I have 6 more tries and it's due at 11:55 pm please help soon if you wish :D

 

[LowlyPion]

I calculated it to be 301.6 m/s

This is the only problem I'm currently missing but I have 6 more tries and it's due at 11:55 pm please help soon if you wish :D

What equation did you use for that. Because you know that it accelerated 33m/s² for 13 seconds. How is that only 301.6m/s?

 

[Eric_meyers]

Oh, I took 33 - 9.8 which is gravity to get my net acceleration for the time intervaql 13 seconds.

 

[Eric_meyers]

PROBLEM SOLVED:

LOL! Thanks, I solved it without subtracting gravity from my acceleration given, that's a poorly worded question though.

I got 99.85% on my physics homework :D I'm at UT Austin physics 1 btw.

 

[LowlyPion]

PROBLEM SOLVED:

LOL! Thanks, I solved it without subtracting gravity from my acceleration given, that's a poorly worded question though.

I got 99.85% on my physics homework :D I'm at UT Austin physics 1 btw.

When they give acceleration that's all you need. When fuel runs out then gravity kicks in.

Good luck.