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.