A child stands on the edge of the cliff 10 m above the ground and throws a stone horizontally with an initial speed of 5 m s$$^{-1}$$. Neglecting the air resistance, the speed with which the stone hits the ground will be _____ m s$$^{-1}$$ (given, $$g = 10$$ m s$$^{-2}$$).

We have a stone thrown horizontally from a cliff of height $$h = 10$$ m with initial horizontal speed $$v_x = 5$$ m/s.

The horizontal velocity remains constant throughout the motion, so $$v_x = 5$$ m/s. For the vertical motion (starting from rest), we use the kinematic equation

$$v_y^2 = u_y^2 + 2gh = 0 + 2 \times 10 \times 10 = 200$$

Now the speed with which the stone hits the ground is

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{25 + 200} = \sqrt{225} = 15 \text{ m/s}$$

Hence, the correct answer is 15 m/s.