# Homework Help: AP free fall question

1. Sep 10, 2008

### jwxie

AP free fall question (solved)

1. The problem statement, all variables and given/known data

I have problem with question b.

A hot air bolloon is traveluing vertically upward at a constant speed of 5.00m/s. When it is 21.0m above the ground, a package is released from the balloon. a) for how long after being released is the package in the air? b) what is the velocity of the package just before impact with the ground?

2. Relevant equations
vf = vi+at
d=vt + 1/2at^2

3. The attempt at a solution

I got Question A with the following

so vf^2 = 5m/s^2 + 2 (-9.8m/s^2)(21m)
and i found the final velocity at the moment it release the package to be 20.9m/s
and i use that as my initial velcoity to find out the answer for question A, and the time is 2.64s

Now, what I don't understand is, since the 2 objects moves at the same speed initially, but if the package is drop, shouldn't the initial velocity stays as 20.9m/s?
i mean i know this is a free-fall question but the correct method is to use 5m/s as my inital velocity (back to the original question), but why?

From my point of view, the balloon carries the package until 21meter, then it releases it, so it descends at 9.8m/s^2. But at the moment when it release, the package is at 20.9m/s upward, so how come we use 5 m/s as our initial in order to find question B???
]
when i tried question B, i use 20.9m/s as my vi instead of 5m/s.
of course, if i use 20.9m/s as my vi, i will get -4.9m/s as my vf which is never possible because the package is going downward as 9.8m/s^2.

the correct answer for question B is -20.9m/s

may someone tell me why?

thank you

Last edited: Sep 10, 2008
2. Sep 10, 2008

### Solaxys

umm I see that as simple mistakes.

1. If they are moving at the same speed up, why would acceleration affect the box? Especially since the balloon is moving at constant speed, which means a = 0.
[You confuse me with that question]

2. you answered it with A.

Again deduce your questions better, I can't undrestand them.

3. Sep 10, 2008

### jwxie

never mind guys
solved it

again, feel so stupid for didn't remember "constant speed"

4. Sep 10, 2008

### jwxie

sorry my friend

i think the approach is somehow right, since 5m/s and 20.9m/s is initial and final as in question a and b
(the mistake is like what you said, constant speed...)

the problem is, i think you and i were right about constant speed
if they are moving at constant speed, then even at the moment the package was released, the speed is at 5m/s