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Old Apr 5th 2009, 07:10 PM   #1
s3a
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Posts: 129
Vertical/Horizontal distance problem

"A balloon is rising at a vertical velocity of 4.9m/s. At the same time, it is drifting horizontally with a velocity of 1.6m/s. If a bottle is released from the balloon when it is 9.8m above the ground, determine (a) the time it takes for the bottle to reach the ground, and (b) the horizontal displacement of the bottle from the balloon."

Here's what I attempted:

/\d vertical = -4.9 m/s^2*(/\t)
-9.8m = -4.9 m/s^2*(/\t)
2s^2 = (/\t)^2
/\t = 1.41s

/\d horizontal = 1.6m/s (1.41s)
/\d horizontal = 2.26m

Here are the answers from the back of the book:

a) 2s
b) 0

PLEASE HELP!!!
Thanks in advance!

Last edited by s3a; Apr 5th 2009 at 07:18 PM.
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Old Apr 8th 2009, 06:30 PM   #2
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Originally Posted by s3a View Post
"A balloon is rising at a vertical velocity of 4.9m/s. At the same time, it is drifting horizontally with a velocity of 1.6m/s. If a bottle is released from the balloon when it is 9.8m above the ground, determine (a) the time it takes for the bottle to reach the ground, and (b) the horizontal displacement of the bottle from the balloon."

Here's what I attempted:

/\d vertical = -4.9 m/s^2*(/\t)
-9.8m = -4.9 m/s^2*(/\t)
2s^2 = (/\t)^2
/\t = 1.41s

/\d horizontal = 1.6m/s (1.41s)
/\d horizontal = 2.26m

Here are the answers from the back of the book:

a) 2s
b) 0

PLEASE HELP!!!
Thanks in advance!
$\displaystyle \Delta y = v_{oy}t - \frac{1}{2}gt^2$

$\displaystyle -9.8 = 4.9t - 4.9t^2$

$\displaystyle -2 = t - t^2$

$\displaystyle t^2 - t - 2 = 0$

$\displaystyle (t - 2)(t+1) = 0$

$\displaystyle t = 2$ sec

as the bottle falls, the balloon stays directly over it ... both have the same horizontal velocity. therefore, horizontal displacement relative to the balloon is 0.
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