Final answer:
The question relates to the physics of kinematics, concerning a bag dropped from a rising helicopter. By using the kinematic equation for displacement with constants for gravity, initial velocity, and time, we calculate the distance fallen as 42.875 m.
Step-by-step explanation:
The subject of this question is physics and it appears to be at the high school level, dealing with kinematics, specifically the free fall motion of dropped objects from a moving vehicle. The situation involves a bag dropped from a rising helicopter, and we are asked to calculate the distance it falls in a given time. To find this distance, we use the kinematic equation for an object's displacement under constant acceleration:
s = ut + (1/2)at²
Where:
- s is the displacement in meters,
- u is the initial velocity in meters per second (m/s),
- a is the acceleration due to gravity (which is 9.8 m/s² in the downward direction), and
- t is the time in seconds.
Given that:
- The initial velocity u is 4.9 m/s (upward), which can be considered as -4.9 m/s in this equation because the positive direction is upward,
- The time t is 2.5 s, and
- The acceleration a is -9.8 m/s² (the negative sign indicates that gravity is in the opposite direction of the initial motion).
Plug these values into the equation to find s:
s = (-4.9 m/s) • (2.5 s) + (1/2) • (-9.8 m/s²) • (2.5 s) ²
s = -12.25 m + (-30.625 m)
s = -42.875 m
The negative sign indicates that the displacement is in the opposite direction of the positive (upward) axis. The absolute value gives the actual distance fallen. Thus, the bag falls 42.875 m.