If a rubber ball falls from a height h and rebounds upto the height of h/2. The percentage loss of total energy of the initial system as well as velocity of ball before it strikes the ground, respectively, are:

A ball is released from a height h. By applying energy conservation, its speed just before striking the ground is given by $$ v = \sqrt{2gh} $$.

After impact, the ball rebounds to a height of h/2. The potential energy at the initial height is $$E_i = mgh$$, while at the rebound height it is $$E_f = mg(h/2) = mgh/2$$.

The energy lost during the collision is therefore $$\Delta E = mgh - mgh/2 = mgh/2$$, which corresponds to a percentage loss of $$\frac{\Delta E}{E_i} \times 100 = \frac{mgh/2}{mgh} \times 100 = 50\%$$.

Hence, the correct answer is Option 1: $$50\%,\ \sqrt{2gh}$$.