To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as


The angular velocity of a body can be described as a function of frequency as



PART A) The expression for the maximum angular velocity is given by the amplitude so that



PART B) The maximum acceleration on your part would be given by the expression



Answer:
leverage
Explanation:
Leverage ...this allows you to use less force through a longer distance
1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration
directed downward, initial vertical position
, and initial vertical velocity
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

substituting, we find

2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of

In order to do that, we use again the same SUVAT equation substituting
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

Which means that the heigth of the packet was

In this case, the movement is uniformly delayed (the final
rapidity is less than the initial rapidity), therefore, the value of the
acceleration will be negative.
1. The following equation is used:
a = (Vf-Vo)/ t
a: acceleration (m/s2)
Vf: final rapidity (m/s)
Vo: initial rapidity (m/s)
t: time (s)
2. Substituting the values in the equation:
a = (5 m/s- 27 m/s)/6.87 s
3. The car's acceleration is:
a= -3.20 m/ s<span>^2</span>