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

Initial horizontal velocity, $u_x = 20 \text{ m/s}$

Initial vertical velocity, $u_y = 0$

Vertical distance, $h = 100 \text{ m}$

Using the equation of motion for the vertical direction:

$v_y^2 = u_y^2 + 2gh$

$v_y^2 = 0 + 2 \times 9.8 \times 100 = 1960$

$v_y = \sqrt{1960} \approx 44.27 \text{ m/s}$

The horizontal velocity remains constant throughout the motion, $v_x = 20 \text{ m/s}$

Magnitude of the final velocity is given by:

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 1960} = \sqrt{400 + 1960} = \sqrt{2360} \approx 48.58 \text{ m/s} \approx 49 \text{ m/s}$

Let $\theta$ be the angle the velocity vector makes with the horizontal:

$\tan \theta = \dfrac{v_y}{v_x} = \dfrac{44.27}{20} \approx 2.21$

\theta = \tan^{-1}(2.21) \approx 66^{}^{{}^{\circ}}

Thus, the ball strikes the ground with a velocity of $49 \text{ m/s}$ at an angle of 66^{}^{{}^{\circ}} with the horizontal.