A hot-air balloon is rising upward with a constant speed of 2.45 m/s. When the balloon is 3.22 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?

Question

I know that average speed equals distance / elapsed time and the question gives the speed and distance so would I set this up as 2.45m/s^ = 3.22m / T ?

Answer 1

You are absolutely on the wrong track.

Compass:
vi=2.45m/s
hi=3.22
hf=0

hf=Hi+vi*t-4.9t^2 solve for t. Notice it is a quadratic.

Answer 2

When compass begins to move its velocity is directed upwards and it moves until the velocity becomes zero.
hₒ =vₒ•t1-g•t1²/2
v=vₒ -gt1.
0 = vₒ -g•t1
t1= vₒ/g =2.45/9.8 =0.25 s.
hₒ =vₒ²/2•g =2.45²/2•9.8 =0.306 m
H =hₒ + h = 0.306+3.22 =3.526 m.
Downward motion
H =g•t2²/2
t2=sqrt(2•H/g) = sqrt (2•3.526/9.8) =0.85s.
t =0.25 +0.85 = 1.1 s.