An Erect Man Falls Into Water from a Platform

[Destructo_Dav]

Homework Statement

A 78kg man, standing erect, steps off a 2.6m high diving platform and begins to fall from rest. The man comes to rest 1.7s after reaching the water.

The acceleration of gravity is 9.8 m/s^2

What average force did the water exert on him?

Answer: 1091.94 N

Homework Equations

F= (deltaP)/(deltaT)
2x=gt^2
v=d/t

The Attempt at a Solution

The first mistake I made was using 1.7s as the time until the man hit the water to find the velocity at the moment before the man hits the water (v = 2.6/1.7), but this got me no where.

I think I need to find the time between when the man jumps and after he falls 2.6m later. I don't know of a way to dervie velocity from only having the distance one falls and the acceleration of gravity. Once I do find that I think I can get the initial velocity and the final velocity, and then I can use the momentum equation to find out the average force exerted. The answer was given to me, but I need to know and understand the process to get it for the test. Anyone have any ideas?

 

[rock.freak667]

At a height of 2.6m, a man of mass 78 kg will have how much gravitational pe? Right before he hits the water, all this energy is converted into kinetic energy. So his velocity before he hits the water is?

 

[Destructo_Dav]

The velocity I found was 7.1386.

Then I set mgh = .5mv^2

78*9.8*2.6 = .5*78*v^2

I found that by v=sqrt(2gh)

Then I used final momentum is equal to initial momentum + (Fnet)(deltaT)

I got 0 = 556.8108 + 1.7Fnet

Fnet = 327.536 N

Where do I go wrong?

 

[Loremaster]

How else can the formula for Newton's Second Law be formulated as opposed to the change in momentum divided by the change in time?

Edit: What I meant was what is the form of the formula most known to people?

 

[Redbelly98]

Fnet = 327.536 N

Where do I go wrong?

You found the net force, but the question asks for the force due to the water. They are different.