Why is the value of kinetic energy always positive?

Vertical component of velocity just after collision $=\frac{u}{2\sqrt[]{2}}$

$k_i=\frac{m{u}^2}{2}{k}_f=\frac{1}{2}m\frac{u^2}{2}+\frac{1}{2}m\frac{u^2}{8}=\frac{5m{u}^2}{16}$ $\frac{}{\text{exception 25:}}$

Fraction  $=\frac{k_i-k_f}{k_i}=1-\frac{\frac{5m{u}^2}{16}}{\frac{m{u}^2}{2}}=1-\frac{5}{8}=\frac{3}{8}$

Using Conservation of Mechanical Energy at point-A and at point-B, we can write
K_B = U_A - U_B [Since K_A = 0]
⇒ (1/2)mv_B² = mg (h_A - h_B)
⇒ v_B = √ (2 × 10 × (10 - 5) = 10m/s

For elastic collision   ${\mathrm{K}\mathrm{E}}_{\mathrm{i}}={\mathrm{K}\mathrm{E}}_{\mathrm{r}}$

$\frac{1}{2}\mathrm{m}\times 25+\frac{1}{2}\times \mathrm{m}\times 9=\frac{1}{2}\mathrm{m}\times 32+\frac{1}{2}m{v}^2$

$34=32+v^2$

$\mathrm{K}\mathrm{E}=\frac{1}{2}\times 0.1\times 2=0.1\mathrm{J}=\frac{1}{10}$

$x=1$

No. Since kinetic energy is a scalar quantity, it only depends on speed of the body and not the direction. So if the direction of the body is changed but the speed remains unchanged, there won't be any effect on the kinetic energy. However, if changing the direction also changes the speed of the body, then kinetic energy of the body will also change.

The 1/2 is a result of mathematical calculation, which occurs when we integrate? vdv in the formula of work done according to Newton's second law of motion. Without this, the final result will turn out to be twice of the actual value.