Posted by **Hannah** on Sunday, May 27, 2012 at 6:45pm.

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?

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 ?

- Physics(thank you for the help) -
**bobpursley**, Sunday, May 27, 2012 at 6:54pm
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.

- Physics(thank you for the help) -
**Elena**, Sunday, May 27, 2012 at 6:56pm
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.

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