Calculating the Force Exerted by Water on a Man Dropping from a Diving Board

Question:

An 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?

Answer:

The force exerted by the water on the man dropping from the diving board is 1626.4 N.

Explanation:

Given Data:
Mass (m) = 82 kg
Distance (S) = 3.0 m
Time (t) = 0.55 s

Calculations:
Since the man starts from rest, the initial velocity (u) is 0 m/s.
Using the second equation of motion: S = ut + (1/2)at^2
Substituting the values: 3 = (1/2) * a * (0.55)^2

Solving for acceleration (a):
3 = (1/2) * a * 0.3025
a = 3 / 0.15125 = 19.83 m/s^2

Calculating the force (F):
F = mass * acceleration
F = 82 * 19.83 = 1626.4 N

Therefore, the water exerts a force of 1626.4 N on the man as he drops from the diving board and comes to rest in the water.