Force exerted by water on a diver

[LOannie234]

A(n) 88 kg man, standing erect, steps o! a
3.6 m high diving platform and begins to fall
from rest. The man comes to rest 2.3 s after
reaching the water.
The acceleration of gravity is 9.8 m/s^2

What average force did the water exert on
him?

My friend and I have tried many different approaches to this but cannot figure it out

 

[tiny-tim]

Welcome to PF!

Hi LOannie234! Welcome to PF!

(try using the X² icon just above the Reply box )

Split it into two parts, air and water, and use the standard constant acceleration equations …

what do you get? ​

 

[LOannie234]

i used the equation vi=vf+at where vi=0 and a=9.81 and t=2.3 and solved for vf which ended up being -22.54. i then plugged that into the equation m(v-vo)/delta t and got the wrong answer

 

[tiny-tim]

i used the equation vi=vf+at where vi=0 and a=9.81 and t=2.3

but that's the equation for falling from rest through air

start again ​

 

[LOannie234]

can you direct me in the right direction? I honestly have no idea how to do this problem. I have been working it for nearly an hour

 

[tiny-tim]

Do the air part first …

find v_f for when the diver hits the water …

what do you get? ​

(that value will then become your v_i for the water part. )

 

[LOannie234]

Using the conservation of energy,
mgh=(1/2)mv^2
so
v=(2gh)^(1/2)

which gets me 8.4m/s

This is then the velocity he enters the water, so

v=vf+at
0=8.4m/s+a(2.3s)
a=-3.652m/s^2

Then F=ma,
F=(88kg)(-3.652m/s^2)=-321.391 N

This is still not close to the answer. Where else am I going wrong?

 

[tiny-tim]

That's the total force … now subtract the force of gravity.

 

[LOannie234]

thanks so much! Moving onto that other problem... where would i start? I have no idea

 

[tiny-tim]

start by starting