# Kinetic/Potential Energy Problem

1. Jul 24, 2013

### mathpat

1. The problem statement, all variables and given/known data

Given
mass = 90kg
distance = 20m above sea level.

Find his potential and kinetic energy at the top of the cliff, at what point in his leap is his kinetic energy 450 J and at what speed does he enter the water?

2. Relevant equations

3. The attempt at a solution

I can solve for PE and KE at the top of the cliff. Have trouble calculating when his KE = 450 J and his velocity entering the water.

2. Jul 24, 2013

### TSny

Hello, mathpat. Could you please state the problem exactly as given to you? It's not clear what's going on at the top of the cliff initially.

3. Jul 24, 2013

### mathpat

sure it's as follows:

Sherlock, a cliff diver, is preparing for his final cliff jump into the ocean. He has a mass of 90 kg and stands on a cliff 20m above the sea. What is his PE and KE at the top of the cliff? At what point in his leap is his KE 450J? At what speed does he enter the water?

4. Jul 24, 2013

### TSny

OK. What did you get for the initial KE and PE at the top of the cliff?

At any point during the fall, what quantity is the same as at the top of the cliff?

5. Jul 24, 2013

### mathpat

I got 17640J for PE. And that would also equal the KE at the top of the cliff?

6. Jul 24, 2013

### TSny

OK for the PE. Why would the KE equal the PE?

7. Jul 24, 2013

### mathpat

Due to no resistance or friction from a free-fall position

8. Jul 24, 2013

### TSny

I don't follow. What determines the KE of an object?

9. Jul 24, 2013

### mathpat

whether the particle or object is moving. Wow ok lol. So the KE = 0 J at the top of the cliff.

10. Jul 24, 2013

### TSny

Yes. Since the problem didn't state whether or not Sherlock jumps from the cliff with an initial velocity, I think you are going to have to assume that he doesn't jump. He just falls from rest.

So, good. His initial KE is zero.

Can you think of a way to find the point where the KE is 450 J?

11. Jul 24, 2013

### mathpat

I'm seriously stumped when it comes to that part. I know I can't plug in that value in KE = 1/2 mv^2 because that would not give me a point.

12. Jul 24, 2013

### TSny

Right, you already know that KE is 450 J at the point you are interested in. So, you don't need to calculate KE there. You are dealing with energy concepts here. Energy is really important due to a very basic principle that you have studied. Can you think of what that principle is?