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

1 Answer
Hammer
Jun 13, 2018

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#

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