A ball is rolled off the edge of a horizon

Vertical component of velocity of ball at point $P$

${v_V} = 0 + gt = 10 \times 0.4 = 4\,m/s$

Horizontal component of velocity = initial velocity

$ \Rightarrow {v_H} = 4\,m/s$

So the speed with which it hits the ground

$v = \sqrt {v_H^2 + v_V^2} = 4\sqrt 2 \,m/s$

and $\tan \theta = \frac{{{v_V}}}{{{v_H}}} = \frac{4}{4} = 1$ $⇒$ $\theta = 45^\circ $

It means the ball hits the ground at an angle of $45^\circ $ to the horizontal.

Height of the table $h = \frac{1}{2}g{t^2} = \frac{1}{2} \times 10 \times {(0.4)^2} = 0.8\,m$

Horizontal distance travelled by the ball from the edge of table $h = ut = 4 \times 0.4 = 1.6\,m$