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
is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)