# Simple Kinematics

1. Jul 15, 2011

### GreenPrint

Even though this is from my computer science textbook, I strongly believe this is a question with regards to physics/calculus. I'm not having an issue with using MATLAB but am stuck on interpreting my results which have to do with calculus/physics. When asked to find the time when the balloon hits the ground I come up with 51.1942 hours. This is a problem because the function that determines displacement is only defined on 0 =< t <= 48, as you'll see when you read the question to the problem. Hence, I don't know how to proceed.

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

Let the following polynomial represent the altitude in meters during the first 48 hours following the launch of a weather balloon:

h(t)=-0.12t^4+12t^3-380t^2+4100t+220

Assume that the units of t are hours.

(a) Use MATLAB together with the fact that the velocity is the first derivative of the altitude to determine the equation for the velocity of the balloon.
(b) Use MATLAB together with he fact that acceleration is the derivative of velocity, or the second derivative of the altitude, to determine the equation for the acceleration of the balloon.
(c) Use MATLAB to determine when the balloon hits the ground. Because h(t) is a fourth-order polynomial, there will be four answers. However, only one answer will be physically meaningful.

There are other parts but part C is were I'm having a problem.

2. Relevant equations

3. The attempt at a solution

Code (Text):
h=sym('-.12*t^4+12*t^3-380*t^2+4100*t+220');
disp('(a)')
V=diff(h,1)
disp('(b)')
a=diff(h,2)
disp('(c)')
max(double(solve(h)))
(a)

V =

- 0.48*t^3 + 36*t^2 - 760*t + 4100

(b)

a =

- 1.44*t^2 + 72*t - 760

(c)

ans =

51.1942

2. Jul 15, 2011

### GreenPrint

I can assume that the answer is about 42.424776102064832588144711559012 sense it's when velocity is equal to zero and afterwards becomes negative or undefined correct?

3. Jul 15, 2011

### GreenPrint

The only problem I'm running into is that h(42.424776102064832588144711559012) is not zero...

I strongly believe that the answer is roughly 51.1942, which is a problem...

Is there an error with my book?

4. Jul 15, 2011

### Dick

If it's going up then velocity is positive. If it's coming down velocity is negative. Why do you think there is something 'undefined' about having negative velocity?? I think your answer of 51.1942 is correct.

5. Jul 15, 2011

### GreenPrint

Well when the velocity is equal to zero it's a critical point of the displacement function and can be a local max or a local minimum of the displacement function. I'm looking for the minimum value of h(t) and I have to test the end points. The function is only defined on [0,48]. The velocity is equal to zero when t =

8.2907656948903308546403410746918
24.284458203044836557214947366296
42.424776102064832588144711559012

and

h(8.2907656948903308546403410746918)=1.4364e+004
h(24.284458203044836557214947366296)=5.8093e+003
h(42.424776102064832588144711559012)=1.7779e+004
h(0)=220
h(48)=1.1594e+004

6. Jul 15, 2011

### GreenPrint

I'm just now realizing that in my code I did max(double(solve(h))) and solved for h and solved for the max value when they wanted me to use velocity to find the answer... still stumped though

7. Jul 15, 2011

### Dick

They want you to find when it hits the ground. That's when h=0. The velocity doesn't have to zero when it hits the ground. You don't need to use the velocity to find that. h is defined everywhere. You just have to find the first point where it becomes 0.

8. Jul 15, 2011

### GreenPrint

Code (Text):
>> solve(h)

ans =

-0.053393860663345759885828826694241
51.194203288621121398687444073266
24.429595286021112180599192376714 - 8.5963040392928577204810707919835*i
8.5963040392928577204810707919835*i + 24.429595286021112180599192376714
I got 51.194203288621121398687444073266 which is out of the range for which h is defined.

9. Jul 15, 2011

### Dick

I think your answer is fine. If they really insist that the formula is only valid for the first 48 hours, then you are right. It's going to be difficult to say where it 'hits the ground'. I think it's a problem with the book.

10. Jul 15, 2011

### GreenPrint

Yay ^_^ ya I think so to