The situation we have here can be visualised below:

First of all, let's define what everything is; #F_f# is the friction force, #F_n# is the normal force, #W# is the weight and #varphi# is the angle at which the ramp is inclined at.

**Note:** The big arrow symbolises movement, not a force.

Now, we usually want to have our forces have the same direction, but in the case of the inclined plane, the weight #W# doesn't really behave nicely, which is why we instead break it down into #W_x# and #W_y#, forces which are parallel to the hypotenuse of the ramp and perpendicular to it, respectively.

These are not random, however. They have the property that

#vec (W_x)+vec(W_y)=vecW#

Also, the angle between #W_y# and #W# is #varphi# due to the similarity of the triangles.

As it isn't stated, we assume that no force is pushing the object.

In order to solve this, we have to use the **variation of Mechanical Energy**. That, or the variation of Kinetic Energy, but I think that that's a bit more complicated.

Let #E# be the mechanical energy, #E_k# be the kinetic energy and #E_p# be the gravitational potential energy.

The variation of mechanical energy, #Delta E#, is:

#Delta E=w_("ext")=E_f-E_i#

where #w# is the work.

What does this mean? Well, it means that if an object moves from point #A# to point #B#, the mechanical energy of is changed by a quantity #DeltaE#, which equal to the work of all the external forces in the system. An external force is one that arises from outside of the system, for example #F_f#.

Let's get into the juicy part.

We have initial velocity #v_i=5m"/"s# and initial height #h_i =0#. When the box reaches the point of maximum height, its velocity is #v_f = 0# and its height #h_("max")#. Using the variation of mechanical energy, we have:

#Delta E = w_("ext") = Fs#, where #s# is the displacement; what we want to find.

Now, the only forces which act on the direction of movement are the forces #F_f# and #W_x#. As they act against it, the work will negative.

#Delta E = -(F_f+W_x)s#

Since the variation of energy is just #E_("final") - E_("initial")#, we have:

#E_f-E_i = -(F_f+W_x)s#

Knowing that the mechanical energy in a point #C# is

#E_c = E_(kC)+E_(pC)#

We can deduce that #E_f# is comprised totally of the potential energy at that point; this is due to the fact that the velocity #v_f#. Similarly, #E_i# is only made up of kinetic energy.

#:. E_(pf)-E_(ci) = -(F_f+W_x)s#

Let's start calculating everything:

#E_(pf) = mgh_("max")#

#E_(ci)=(mv_i^2)/2=(25m)/2#

#F_f = F_nmu#

Where #mu# is the coefficient of friction. At the same time, #F_n = W_y# due to the fact they are opposite and that the box is not moving vertically (perpendicular to the ramp, that is)

#F_f = W_y*1/2#

In our triangle, we can get the following results:

#sin varphi=W_x"/"W=> W_x=Wsinvarphi#

#cosvarphi = W_y"/"W => W_y=Wcosvarphi#

As #varphi = (2pi)/3#:

#W_x=sqrt3/2W and W_y=1/2W#

#=> F_f = 1/4W=1/4mg#

About #h_max#, we can get that #h_max=ssinvarphi#

Hence, we get:

#cancelm(sqrt3/2gs-25/2)=-cancelm(1/4g+sqrt3/2g)s#

We can see that the mass #m# doesn't matter, as it is simplified.

Now, get everything featuring #g# and #s# on the left hand side:

#sqrt3/2gs+1/4gs+sqrt3/2gs=25/2#

#gs(sqrt3+1/4)=25/2#

#gs(4sqrt3+1)=50#

#color(red)( :.s=50/(g(4sqrt3+1)) ~~ 0.6306#