1- interaction between 2 objects
2- action- reaction force pairs
Answer:
1626.4 N
Explanation:
Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?
The parameters to be considered are:
Distance S = 3m
Time t = 0.55s
Since the man started from rest, initial velocity u = 0
Using second equation of motion
S = Ut + 1/2at^2
3 = 1/2 × a × 0.55^2
3 = 1/2 × a × 0.3025
a = 3/ 0.15125
a = 19.83 m/s^2
Force = mass × acceleration
Force = 82 × 19.83
Force = 1626.4 N
Therefore, the force that water exerted on him is 1626.4 N
Answer:
Displacement from the starting position is 103.21m
Explanation:
If you draw these directions, it will create the two legs of a triangle.
Using this method, you can visualize why your displacement is what it is.
Using the pythagorean theorem

Plug in both values


c = 103.2085
c= 103.21
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brainly.com/question/24937533
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