Question

At what height would the Kinetic energy of a falling particle be equal to half of its potential energy?

Answer 1

If the particle starts falling from height `H` and with initial speed `v_0`, then the wanted height is

`h=1/3(2H+v_0^2/g)`

Explanation:

See below.

If we denote `K` and `U` to be the kinetic and potential energy respectively, we have to solve the equation

`K=1/2U`

Knowing that `K = (mv^2)/2` and `U=mgh`, where `m` is the mass of the particle, `v` is the speed at that certain height and `h` is that height:

`(cancelmv^2)/cancel2 = (cancelmgh)/cancel2`

`v^2=gh => h=v^2"/"g`

Of course, we still don't know the speed `v`. However, we can write it in terms of the initial speed, `v_0`.

Galilei's formula, applied for a falling particle, states that

`v_A^2=v_0^2+2gd`

Where `v_A` is the speed at some point `A` and `d` is the distance traveled from the initial falling point from the current position, `A`. As such, `d = h_("total") - h_(A)`.

As `h_A` is the height we want to find,

`v^2=v_0^2+2g(h_"total"-h)`

Let `h_"total"` be `H`.

`v^2=v_0^2+2g(H-h)`

`:. h= (v_0^2+2g(H-h))/g=v_0^2/g+2H-2h`

`:. 3h=v_0^2/g +2H=> color(red)(h=1/3(2H+v_0^2/g))`

If the object is freefalling, a.k.a. `v_0=0`, the height is

`h=2/3H`