Answer of your question is in this photo
In this question, you're determining the time (t) taken for an object to fall from a distance (d).
The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)
d = 8,848m and g = 9.8m/s
Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)
The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)
Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)
Now for the last step, find the square root of the remaining number:
t = 42.5s
So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.
I hope this helps :)
What are the two sets? I only see the two answers?
Answer:
B) collision is inelastic because they stick together after collision and share a common final velocity Vf
C) M1V1 + M2V2 = (M1 + M2)Vf
D) Vf = 6.33m/s
E) force = 3040N
Explanation:
Detailed explanation and calculation is shown in the image below
To solve the problem it is necessary to apply conservation of the moment and conservation of energy.
By conservation of the moment we know that

Where
M=Heavier mass
V = Velocity of heavier mass
m = lighter mass
v = velocity of lighter mass
That equation in function of the velocity of heavier mass is

Also we have that 
On the other hand we have from law of conservation of energy that

Where,
W_f = Work made by friction
KE = Kinetic Force
Applying this equation in heavier object.






Here we can apply the law of conservation of energy for light mass, then

Replacing the value of 

Deleting constants,

