rotational mechanics pls try this

let the ball fall on the floor with final velocity (-v).
e=velocity of recoilvelocity of approach=1
so,velocity of recoil=velocity of approach

so in vertical direction,
velocity of recoil = v

now let impulsive normal reaction N act during collision,
thus Ndt=change in momentum=mv-(-mv)=2mv....(i)

now there wud be impulsive friction acting on the sphere due to impulsive normal reaction,
f=μN.

(Impulsive torque).dt=change in angular momentum=Iω-Iω_o
or, μNRdt=Iω-Iω_o
or, Iω=μNRdt+Iω_o

now impulsive impulse will give horizontal impulse..
so, μNdt=mv_horizontal
or, μ(2mv)=mv_horizontal [from (i)]
or, v_horizontal=2μv

now there will be projectile motion in which
v_y=v
and v_x=2μv

so the range of this projectile will give us the distance between points of first and second impact of the ball with the floor.

and for part two we will have to do this

the energy in vertical direction is conserved....

so loss in energy=12Iω_o²-12Iω²+12m(2μv)²-

the answer will be thus achieved!!![1][1][1]

can anyone give a gym-ed solution???

my solution is just so MOTA!!![2][2]

i am asking for a slimmer solution!! nishant bhaiya can u help?

som1 plz try! i ll check out after waking up!! bye bye!!