Answer:
The average velocity is
and
respectively.
Explanation:
Let's start writing the vertical position equation :

Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt = 
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :

For the position variation we use the vertical position equation :

Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is

For the second time interval :
t1 = 4 s → t2 = 9 s


Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :

Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then


The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is

Answer:
Explanation:
The formula for the magnitude of a vector is
and then round to the hundredths place:
3.11 m. Since we are in Q2, we can also find the direction of this vector:
but since we are in Q2, we add 180 degrees to the result, getting the angle to be 115.3
Answer:
27.1m/s
Explanation:
Given parameters:
Height of the building = 30m
Initial velocity = 12m/s
Unknown:
Final velocity = ?
Solution:
We apply one of the kinematics equation to solve this problem:
v² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 12² + (2 x 9.8 x 30)
v = 27.1m/s
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