How Fast Does a Swimmer Hit the Water from 10 Meters?

[TNewC]

Homework Statement

A 65kg swimmer jumps of a 10m tower.
a) Find the swimmer's velocity when hitting the water
b) The swimmer comes to a stop 2.0m below the surface. Find the net force exerted by the water.

Homework Equations

F=ma?
I don't know the rest, which is part of the problem

The Attempt at a Solution

I'm entirely lost, I can't figure out how to find velocity when you are given mass and distance. I just don't understand. I only got as far as being able to state the data. Please help!

 

[denverdoc]

well, deep breath, quiet the mind, and steel your resolve.

What force is operating on the diver and what does that do his speed? Forget the numbers for a minute.

 

[TNewC]

The force of gravity is acting on the diver, and wouldn't that increase his speed?

 

[denverdoc]

yes, good. Do you have a number for g, the acceleration due to gravity?

 

[TNewC]

Yes, 9.8m/s²

 

[denverdoc]

Perfect, now what that constant tells you is for every second, the speed will increase by 9.8m/s. So At 1 sec v=9.8m/s at 4 sec=4*9.8m/s going too slow?

 

[TNewC]

Okay, that makes sense, but I don't have a time.

 

[denverdoc]

Bingo. But we are given height, not the time he falls. But there is some help. By using any number of approaches we find that for anybody undergoing constant acceleration or deceleration as he is about to be:

2*a*x=V'^2-V^2 where d is the distance traveled while under the influence of the force.

Edit sorry, not getting the latex to work right v' is the final velocity, v the initial velocity

 

[denverdoc]

In this case as he "falls" off the diving board, the initial velocity is zero. So you now have a way using distance to calculate V.

 

[TNewC]

Oh! Could I use d=1/2at² to calculate time?

 

[denverdoc]

you could, and multiply g times the t you get. This will work, the eqn above is streamlined.

 

[TNewC]

Okay, thank you so much! I now understand what I need to do for part a, but I still don't quite get part b...can you help me with that too?

 

[denverdoc]

sure, the process is exactly like what we just did. Only now we have to slow the diver through the depth of the water he reaches which is the displacement in this case. We know tyhe speed as he enters the water. We could do the time thing as before, but my advice would be two use tse the eqn:

a(2*d)=v^2 where d is 2 meters. Then of course F=ma allows you to complete (b)

 

[TNewC]

Thank you so much! You have quite possibly saved my Physics grade. I understand this now!

 

[denverdoc]

Thank you so much! You have quite possibly saved my Physics grade. I understand this now!

Life was tougher before the net...I damn well recall how greek this all seemed once upon a time.