From a point 100 m above the ground, a ball is thrown horizontally...

Let the time taken by the ball to reach the ground be $t$.

Using the second equation of motion in the vertical direction:

$h = u_y t + \dfrac{1}{2} g t^2$

Since the ball is thrown horizontally, the initial vertical velocity $u_y = 0$. Assuming $g = 10 \text{ m/s}^2$:

$100 = 0 + \dfrac{1}{2} (10) t^2$

$100 = 5t^2$

$t^2 = 20 \Rightarrow t = \sqrt{20} = 2\sqrt{5} \text{ s}$

The horizontal distance (range) traveled by the ball is given by:

$R = u_x \times t$

$R = 20 \times 2\sqrt{5} = 40\sqrt{5} \text{ m}$