A ball is thrown horizontally from a cliff 50m high with an initial velocity of 10m/s. Calculate the time to hit the ground, final vertical velocity, and horizontal range.

The vertical displacement is $H$. Since the initial vertical velocity is zero, the equation is:

$$H = u_y t + \frac{1}{2}gt^2$$

$$50 \, \text{m} = (0 \, \text{m/s})t + \frac{1}{2}(10 \, \text{m/s}^2)t^2$$ $$50 = 5t^2$$

$$t^2 = \frac{50}{5}$$ $$t^2 = 10$$ $$t = \sqrt{10} \, \text{s}$$ $$t \approx 3.162 \, \text{s}$$

The time taken to reach the ground is $\boxed{\text{3.16 s}}$.

$$v_y = 0 \, \text{m/s} + (10 \, \text{m/s}^2)(\sqrt{10} \, \text{s})$$ $$v_y = 10\sqrt{10} \, \text{m/s}$$ $$v_y \approx 31.62 \, \text{m/s}$$

The vertical component of the velocity when the ball hits the ground is $\boxed{\text{31.6 m/s}}$.

Since there is no horizontal acceleration, the horizontal velocity remains constant.

$$R = u_x t$$

$$R = (10 \, \text{m/s})(\sqrt{10} \, \text{s})$$ $$R = 10\sqrt{10} \, \text{m}$$ $$R \approx 31.62 \, \text{m}$$

The distance from the foot of the cliff to where the ball hits the ground is $\boxed{\text{31.6 m}}$.