Solving for potential energy after time t.

[blipped]

Homework Statement

A stone of mass 0.2kg is dropped from the top of a building 80m high. After t seconds it has fallen x meters and has a velocity of v. What is the potential energy of the stone t seconds later

Homework Equations

The answer is 2(80 - x) = 160 - 10t^2

Using suvat: s = ut + (1/2)at^2

The Attempt at a Solution

I started with PE = mgh
PE = 0.2 * 10 * h
PE = 2h

Then I figured h = (1/2)*10*t^2
or h = 5t^2

So now I have

PE = 10t^2

Not sure if I'm headed in the right direction or not.

 

[cnh1995]

So now I have

PE = 10t^2

That gives PE=0 at the instant of dropping. Where should PE be 0 and where should it be maximum?

 

[cnh1995]

A stone of mass 0.2kg is dropped from the top of a building 80m high.

It is coming downwards.

s = ut + (1/2)at^2

What is s?

PE = mgh

What is h?
Are they both same in this problem?

 

[blipped]

That gives PE=0 at the instant of dropping. Where should PE be 0 and where should it be maximum?

I got it, or at least I understand the answer now. 80 - (80 - x) gives the remaining distance with regards to x, and this equal to the same measurement given (1/2)(g)t^2. Set them equal and shuffle around a bit. Thanks

 

[cnh1995]

I got it, or at least I understand the answer now. 80 - (80 - x) gives the remaining distance with regards to x, and this equal to the same measurement given (1/2)(g)t^2. Set them equal and shuffle around a bit. Thanks

Right.
x=80-h.