A small sandbag is dropped from rest from a hovering hot air balloon. After 2.0 s, how far below the balloon is the sand bag?
use
S=ut+(1/2)at²
S=displacement, m
u=initial velocity = 0 m/s
t=time = 2 s.
a=acceleration = acceleration due to gravity = -9.8 m/s/s
Well, if it's a small sandbag, we can assume it's not a heavy hitter. So, after 2.0 seconds, I'm guessing the sandbag has fallen a few feet below the balloon. Just enough to keep things interesting, you know?
To determine how far below the hot air balloon the sandbag is after 2.0 seconds, we can use the equation of motion for freefall:
y = (1/2)gt^2
Where:
- y represents the vertical distance traveled by the sandbag,
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- and t is the time that has elapsed (2.0 seconds in this case).
Substituting the given values into the equation, we can find the answer:
y = (1/2)(9.8 m/s^2)(2.0 s)^2
y = (1/2)(9.8 m/s^2)(4.0 s^2)
y = (4.9 m/s^2)(4.0 s^2)
y = 19.6 m
Therefore, after 2.0 seconds, the sandbag will be located 19.6 meters below the hot air balloon.