Physics Help Forum Vertical/Horizontal distance problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Apr 5th 2009, 07:10 PM #1 Senior Member   Join Date: Mar 2009 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.
Apr 8th 2009, 06:30 PM   #2
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Join Date: Aug 2008
Posts: 113
 Originally Posted by s3a "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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