Can changing the direction of a body also affect it's kinetic energy?

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   ${\mathrm{K}\mathrm{E}}_{\mathrm{i}}={\mathrm{K}\mathrm{E}}_{\mathrm{r}}$

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

$34=32+v^2$

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

$x=1$

If you look closely at the formula of kinetic energy (1/2*m*v^2), the velocity is squared which automatically gives a positive integer. And mass of the body can never be a negative value, which leads to the result being a positive integer.

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.