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
The first two are always the reactants the products come after so they are last
Answer:
Equation for SHM can be written
V = w A cos w t where w is the angular frequency and the velocity is a maximum at t = 0
V1 = w1 A cos w1 t
V2 = w2 A cos w2 t
V2 / V1 = w2 / w1 since cos X t = 1 if t = zero
V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2
If the velocity is twice as large the period will be 1/2 long
Mountains, tops of buildings, and high-flying aircraft are all part of Earth's atmosphere, no matter how high they are. On the other hand, space doesn't belong to our atmosphere, it is outside of it. Having this in mind, the best location to place a telescope used to observe x-rays from stars is in space.
The correct answer is C) Rainfall
