# One Dimensional Hot Air Balloon Kinematics

1. Sep 13, 2009

### Flippit

1. The problem statement, all variables and given/known data
A hot-air balloon has just lifted off and is rising at the constant rate of 2.3 m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 12 m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her?

2. Relevant equations

One-Dimensional Kinematics

3. The attempt at a solution

First I tried setting the distances of the balloon and the camera equal to each other,

camera: 12t+.5*-9.81t^2
balloon: 2.3t+2.5

12t+.5*-9.81t^2=2.3t+2.5

then I rearranged the equation

12t-4.905t^2-2.3t=2.5
9.7t-4.905t^2-2.5=0
-4.905t^2+9.7t-2.5=0

Next, I plugged the equation into the quadratic formula.

t= -9.7+-sqrt 9.7^2-4.905*-2.5/2*-4.905
t= -9.7+-sqrt 94.09+19.62*-2.5/-9.81
t= -9.7+-sqrt 94.09-49.05/-9.81
t= -9.7+-sqrt 45.04/9.81
t= -9.7+-6.71/9.81
t= .305, 1.67

next, I plugged the t value into the formula x= Xi + Vi*t + .5at^2
x= 0 + 12*.305 + .5*-9.81*.305^2
x= 3.66-4.905*.093
x= 3.66 -.456
x= 3.2m

I entered this into the website we turn our homework into, but it said the answer was incorrect, I'm not really sure what I did wrong, all my math seems correct.

Last edited: Sep 13, 2009
2. Sep 13, 2009

### tiny-tim

Welcome to PF!

Hi Flippit! Welcome to PF!

(try using the X2 tag just above the Reply box )
Looks ok to me.

(except I would have used the 2.3t+2.5 equation at the end, instead of the quadratic )

3. Sep 13, 2009

### Flippit

Alright, I guess I'll just have to talk to my teacher about it. Thanks for looking it over!