# Basic Kinematics Balloon

1. Feb 13, 2008

### gmunoz18

Ive been stressing on this problem for a while ill show you the work below

A hot-air balloon is rising upward with a constant speed of 2.60 m/s. When the balloon is 3.20 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

so first i did the time it took for the balloon to reach 3.2 meters which was 3.2/2.6= 1.23 seconds

and than i did the falling object which i got .808 seconds. using -9.8 as acceleration and found used the basic kinematic equations to get the .808 seconds

i added these together for the round trip to hit the ground and got 2.0389 seconds but this number is not right i dont know what im doing wrong. Ive tried just the regular fall from the balloon and it didnt work. im lost confused why it isnt working and frustrated thanks alot in advance

2. Feb 13, 2008

### K.J.Healey

Remember the compass, when dropped, had an initial upward velocity, it didnt drop from rest. Did you take that into account?

3. Feb 13, 2008

### gmunoz18

no i did not im gonna try and work that right now

4. Feb 13, 2008

### K.J.Healey

If you don't get it, I think my answer came out to be t=1.11586 seconds from the time the compass is let go till it hits the ground, so that plus whatever time to get the balloon into the air.

using
Xf-Xi = Vi*t + (1/2) a t^2

Solving for t

5. Feb 14, 2008

### perucho

You have the altitude h and the initial speed V_0, so you may apply
V=V_0-gt_1,
and
V^2=V_0^2-2.g.h_1,
and set V=0 to know the time t_1 and the height h_1 that the compass rise further (up to the max height under the action of gravity). That is,
t_1=V_0/g, and h_1=(V_0)^2/(2g).
So by applying now free fall you have
h_1+h=1/2.g.(t_2)^2 (recall that V=0 there, at the max height)
Solve for t_2 and the answer is
t=t_1+t_2.
That is an instructive way to solve it because you know what are doing.