Final answer:
The bag falls 2.775 meters below the helicopter after being dropped and falling for 1.5 seconds, calculated by taking into account the initial upward velocity and the acceleration due to gravity in opposite directions.
Step-by-step explanation:
To determine how far below the helicopter the bag is after being dropped and falling for 1.5 seconds, we can use the kinematic equations of motion for an object under uniform acceleration. In this scenario, the initial velocity of the bag is 5.5 m/s upward (since the helicopter is rising at this velocity when the bag is dropped), the acceleration due to gravity is 9.8 m/s2 downward, and the time the bag falls is 1.5 seconds.
First, we calculate the displacement due to the initial velocity using the formula s = ut, where s is the displacement, u is the initial velocity, and t is time. For the bag dropped from the helicopter, this gives us a displacement of 5.5 m/s × 1.5 s = 8.25 m in the upward direction. Then, we calculate the displacement due to gravity using the formula s = 0.5at2, where a is the acceleration and t is time. This gives us a displacement due to gravity of 0.5 × 9.8 m/s2 × (1.5 s)2 = 11.025 m in the downward direction.
The net displacement is the difference between these two displacements since they are in opposite directions. Therefore, the bag falls 11.025 m - 8.25 m = 2.775 m below the helicopter after 1.5 seconds.