Hep me solve this Physics

A 5 kg block is projected vertically upward at 20 m/sec from grounds level. The block loses 40% of its initial energy to air resistance as it rises to a maximum height. To what height (in meters) above ground does the block (eventually) rise?

Let the initial energy is E
At maximum height the energy lost is 40% of E = 0.4E
now the energy at max height = 0.6E [energy remaining = (E- 0.4E)]

the energy at max height(final energy of block) is potential energy P.E = mgh
the energy at point of projection(initial energy of block) is kinetic energy K.E = mv^2/2

now given that (final energy of block) = 0.6*(initail energy of block)

=> mgh = 0.6*(mv^2)/2

=> gh = 0.3*v^2

=> h = 0.3*(20)^2/9.8

=> h = 12.24 metres

the max height is 12.24 metres


note: here law of conservation of energy holds...no energy is destroyed

ie., initial energy = final energy

=> initial energy of block = final energy of block + energy lost to air molecules

take ground as reference
Initial Energy = (1/2)5*(20)^2
lost energy = (0.4)* initial energy.
now the rest of the energy will get converted into potential energy.
hence, mgH= left energy. (m=5kg , g=10m/s2, H height it reached)
after above calculations H will come out to be 12 meters.( will be 12.24 if u take g= 9.8m/s2)

Had there been no losses, the height could be found using relation,
h=Vinit^2-Vfinal^2/2*g
As Vfinal=0, h=(20)^2/2*9.8=200/9.8= 20.41m
Since 40% is lost and h is directly proportional to energy, it will be 60% of total height.
Hence, actual height h1=0.6*(20.41)= 12.245 m

mgh = 40% m v² => h = 0.40 v²/g = 0.40 400 / 9.81 = 16.3 m. Note that you don't need the mass.