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jwxie
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AP free fall question (solved)
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?
vf = vi+at
vf^2 = vi^2+2ad
d=vt + 1/2at^2
I got Question A with the following
vf^2 = vi^2 + 2ad
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
Homework Statement
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?
Homework Equations
vf = vi+at
vf^2 = vi^2+2ad
d=vt + 1/2at^2
The Attempt at a Solution
I got Question A with the following
vf^2 = vi^2 + 2ad
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
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