# Quick question on position

1. Sep 20, 2007

### rocomath

a hot air balloonist, rising vertically with a constant velocity of magnitude 5m/s releases a sandbag at an instant when the balloon is 40m above the ground. after it is released, the sandbag is in free fall.

a) compute position at .250s

i solved it and got ~ 40.9m, which is correct. but why is it +.9m above when it was dropped when the position of the balloonist was 40m? shouldn't it be -.9m?

is it b/c even though the sandbag was dropped at the instant the balloonist was at 40m, the position increased due to the velocity which includes a rise in position within that .250s that they were asking for?

Last edited: Sep 20, 2007
2. Sep 20, 2007

### dynamicsolo

To one part of your questions, the answer is: yes, you have to take the initial velocity of the sandbag as 5 m/sec upward because it was moving with the balloon just before it was released.

As for part (a), *what* is it you are asked to compute the position of? You should show some of your work, because I am mystified as to what 40.9 meters represents...