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

Answer: Last option
2.27 m/s2
Explanation:
As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.
If we call a_c to the centripetal acceleration then, by definition

in this case we know the speed of the runner

The radius "r" will be the distance from the runner to the center of the track



The answer is the last option
Answer:

Explanation:
From the question we are told that:
Acceleration 
Displacement 
Initial time 
Final Time 
Generally the equation for Velocity of 1.05 travel is mathematically given by
Using Newton's Law of Motion



Generally the equation for Distance traveled before stop is mathematically given by



Generally the equation for Distance to stop is mathematically given by
Since For this Final section
Final velocity 
Initial velocity 
Therefore
Using Newton's Law of Motion


Giving

Therefore



Generally the Total Distance Traveled is mathematically given by


