Posted by Heart on Wednesday, December 23, 2009 at 9:42pm.
For the velocity, v, on contact with water, we can equate kinetic and potential energies while ignoring air-resistance:
h=10 m
g=9.8 m/s/s
(1/2)mv²=mgh
Solve for v.
For the force, it would be similar to the car acceleration problem, except that to find the deceleration in water, we would use
v1²-v²=2aS
S=distance (2 m)
v1=final velocity =0
v=initial velocity as found above
a=deceleration due to water (negative) including effects of buoyancy and gravity.
and finally
F=ma
What they call the "net force" should be the "average force". The net force changes with time during deceleration. At the low point of the dive, it equals the buoyancy, but it starts out much higher at the time of impact with the water.
A quicker way to get the average force, F, on the diver under water would be to equate the work done against the water to the initial potential energy before the jump. This P. E. should include the potential energy decrease while in the water. Thus:
M g *(10m + 2m) = F * (2 m)
F = (M g)*12/2 = 6 M g
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