MatRobert said:
Working out gravitational energy = mg = (65)*10 = 650
Equating to kinetic energy, (1/2)*(65)*v^2 = 650
v=4.47m/s.
if you add in air resistance.. that velocity just might drop a little.
What you need to do now is calculate the rate at which you displace the volume in water. This might be a little complicated as when you'll dive, you'll change your pose very quickly just as you enter the surface of the water and this rate of change will be very complex.
Once you are in the water, any position change will not result in a change in the force upon you. It will just differ in the force applied to your different body parts. Since we're talking about safety here, i can say that you should see to it that minimum force is applied on the chest.. to protect your lungs and heart.
Let's say.. you completely enter water without changing your position. If you jump in water something like this:
http://img156.imageshack.us/img156/4521/divingwithlennondb5.jpg
Sorry for adding that image.. that's all i could find. Plus.. i did change Yoko Ono to Jessica Alba [ \textrm{Jessica~Alba} \gg \textrm{Yoko~Ono} ].
You are basically a standard ellipsoid where the x-y plane equation is given by:
<br />
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1<br />
You can use the image to find out the approximate eccentricity [I got \varepsilon = 0.896]. Then take a measurement of your height from the head to the hip. This will give you half the circumference of your 'ellipse'. Then using Ramanujan's approximation:
C \approx \pi a \left[ 3 (1+\sqrt{1-\varepsilon^2}) - \sqrt{(3+ \sqrt{1-\varepsilon^2})(1+3 \sqrt{1-\varepsilon^2})} \right]
you can find out 'a' and 'b' for your 'ellipse'
The forces acting are: i] The force due to gravity, ii] The buoyant force, iii] The drag force [fluid resistance].
The force due to gravity will always be constant. The buoyant force will be given by the amount of water displaced. In the image, i have taken an instant where the body is immersed such that the point just above the surface of the water is 'x' distance from the center of the ellipse.
The volume of that part of the ellipse is given by:
<br />
V = \pi b^2 \left(1 - \frac{a}{3} - \frac{x^3}{3a^2}\right)<br />
I found out this using integration. I'm not doing the proof here now.. but it's easy to find out.
Now, the buoyant force will be given by the weight of fluid displaced:
<br />
F_b = - \pi b^2 \left(1 - \frac{a}{3} - \frac{x^3}{3a^2}\right) \rho_w<br />
where \rho_w is the density of water.
Now, the drag force is given by:
<br />
F_d = - \sigma v = - \sigma \frac{dx}{dt}<br />
In my sign convention, all upward forces are negative hence I've taken negative for both F_d and F_b.
Let the acceleration of the body be 'a'. Then,
<br />
F_d + F_b + F_g = ma<br />
giving us:
<br />
- \sigma \frac{dx}{dt} - \pi b^2 \left(1 - \frac{a}{3} - \frac{x^3}{3a^2}\right) \rho_w + mg = m \frac{d^2x}{dt^2}<br />
In this equation '\sigma' is a property of the object that travels through a fluid. You can approximate \sigma = 6 \pi \eta rv.. but that holds good only for spheres. Try to find the \sigma for ellipsoids.
Solving this differential equation (which I'm not able to) will give you x(t). Once you have x(t), you can use that and it's derivatives to find the impulse imparted [i'm guessing that even that is going to be time dependent]. This will help you find the pressure that applies on your body as a function of time. You'll need the area of the ellipsoid to calculate the pressure. You can find that at wikipedia. Do remember that the area exposed to the water will be different at different time intervals.
'Safety data' for your muscles and bones is provided in terms of the yield strength i.e. max. pressure they can tolerate. Once you have the data, you can use this data to find out the safety parameters.
