1. The bag's velocity immediately before hitting the ground.
Recall this kinematics equation:
Vf = Vi + aΔt
Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.
Given values:
Vi = 0m/s (you assume this because the bag is dropped, so it falls starting from rest)
a is 9.81m/s² (this is the near-constant acceleration of objects near the surface of the earth)
Δt = 2s
Plug in the values and solve for Vf:
Vf = 0 + 9.81×2
Vf = 19.62m/s
2. The height of the helicopter.
Recall this other kinematics equation:
d = ViΔt + 0.5aΔt²
d is the distance traveled by the object, Vi is the initial velocity, a is the acceleration, and Δt is the time elapsed.
Given values:
Vi = 0m/s (bag is dropped starting from rest)
a = 9.81m/s² (acceleration due to gravity of the earth)
Δt = 2s
Plug in the values and solve for d:
d = 0×2 + 0.5×9.81×2²
d = 19.62m
3. Time of the bag's fall and its velocity immediately before hitting the ground... if it started falling at 2m/s
Reuse the equation from question 2:
d = ViΔt + 0.5aΔt²
Given values:
d = 19.6m (height of the helicopter obtained from question 2)
Vi = 2m/s
a = 9.81m/s² (acceleration due to earth's gravity)
Plug in the values and solve for Δt:
19.6 = 2Δt + 0.5×9.81Δt²
4.91Δt² + 2Δt - 19.6 = 0
Use the quadratic formula to get values of Δt (a quick Google search will give you the formula and how to use it to solve for unknown values):
Δt = 1.8s, Δt = −2.2s
The formula gives us 2 possible answers for Δt but within the situation of our problem, only the positive value makes sense. Reject the negative value.
Δt = 1.8s
Now we can use this new value of Δt to get the velocity before hitting the ground:
Vf = Vi + aΔt
Given values:
Vi = 2m/s
a = 9.81m/s²
Δt = 1.8s (result from previous question)
Plug in the values and solve for Vf:
Vf = 2 + 9.81×1.8
Vf = 19.66m/s
