Posted by **Heart** on Wednesday, December 23, 2009 at 9:42pm.

A 65 kg diver jumps off of a 10.0 m tower.

a. Find the diver's velocity when he hits the water.

b. The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water.

Ok, im a little confused about this problem. We knoe the mass of the diver is 65 kg and the distance is 10 m.To determine the velocity wouldn't you need the time too?

- physics -
**MathMate**, Thursday, December 24, 2009 at 7:50am
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

- physics -
**drwls**, Thursday, December 24, 2009 at 8:11am
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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