Above Board
2010
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A 97 kg man drops from rest on a diving board 4.0 m above the surface of the water and comes to rest 0.77 sec?
A 97 kg man drops from rest on a diving board 4.0 m above the surface of the water and comes to rest 0.77 seconds after reaching the water. What force does the water exert on him?
Answer --> The force on the man is -1,116 N. Note the negative sign. Force is a vector quantity, meaning that it has a value and a direction. If you leave off the negative sign, the answer is incorrect. By the way, that man weighs nearly 215 pounds. Hefty.
Given the values:
m = 97 kg
d = 4.0 m
t = 0.77 s
Vi = 0.0 m/s
g = 9.81 m/s^2
Using uniform acceleration equations, first find his velocity upon hitting the water:
Vf = SQRT { Vi^2 + [2gd] }
Vf = SQRT { (0.0 m/s)^2 + [ 2 * (9.81 m/s^2) * (4.0 m) ] }
Vf = SQRT { 78.48 m^2/s^2 }
Vf = 8.86 m/s
Find his acceleration coming to halt, using Vf above now equals Vi for this equation
a = [Vf -Vi] / t
a = [ (0.0 m/s) - (8.86 m/s) ] / (0.77s)
a = [ -8.86 m/s ] / (0.77 s)
a = -11.5 m/s^2
Find the force on the idiot from the water from Newton's 2nd Law
F = m * a
F = (97 kg) * (-11.5 m/s^2)
F = -1,116 N
Lowkey - Sorry
