Basic Kinematics Balloon

[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 don't know what I am doing wrong. I've tried just the regular fall from the balloon and it didnt work. I am lost confused why it isn't working and frustrated thanks a lot in advance

 

[K.J.Healey]

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

 

[gmunoz18]

no i did not I am going to try and work that right now

 

[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

 

[perucho]

no i did not I am going to try and work that right now

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.