Answer: Option A
:
A

Vertical component of velocity of ball at point P
$v_v$ = 0 + gt = 10 $\times$ 0.4 = 4 m/s
Horizontal component of velocity = inital velocity
$\Rightarrow$ $v_H$ = 4 m/s
So the speed with which it hits the ground
v = $\sqrt{({v_H}^2+{v_y}^2)}$ = 4$\sqrt{2}$ m/s
and tan $\theta$= $\frac{v_V}{v_H}$ =$\frac{4}{4}$ = 1 $\Rightarrow$ $\theta$ = ${45}^{\circ }$
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 $\times$ ${\left(0.4\right)}^2$ = 0.8m
Horizontal distance travelled by the ball from the edge of table h = ut = 4 $\times$ 0.4 = 1.6m