A body is dropped on ground from a height $$h_1$$ and after hitting the ground, it rebounds to a height $$h_2$$. If the ratio of velocities of the body just before and after hitting ground is 4, then percentage loss in kinetic energy of the body is $$\dfrac{x}{4}$$. The value of $$x$$ is ______.

A body is dropped from height $$h_1$$ and rebounds to height $$h_2$$. The ratio of velocities just before and after hitting the ground is $$4$$.

The velocity just before hitting the ground is $$v_1 = \sqrt{2gh_1}$$ and the velocity just after rebounding is $$v_2 = \sqrt{2gh_2}$$, satisfying $$\dfrac{v_1}{v_2} = 4$$.

Squaring gives $$\dfrac{v_1^2}{v_2^2} = 16$$, so the ratio of kinetic energies after and before impact is $$\dfrac{KE_{\text{after}}}{KE_{\text{before}}} = \dfrac{v_2^2}{v_1^2} = \dfrac{1}{16}$$.

The percentage loss in kinetic energy is $$\% \text{ loss} = \left(1 - \dfrac{v_2^2}{v_1^2}\right)\times 100 = \left(1 - \dfrac{1}{16}\right)\times 100 = \dfrac{15}{16}\times 100 = 93.75\%$$.

Since $$\dfrac{x}{4} = 93.75$$, it follows that $$x = 93.75 \times 4 = 375$$.

The value of $$x$$ is $$\boxed{375}$$.