d
(d) 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 \(\tan \theta = \frac{{{v_V}}}{{{v_H}}} = \frac{4}{4} = 1\)

 \( \Rightarrow\) \(\theta = 45^\circ \)
It means the ball hits the ground at an angle of \(45^\circ \) to the horizontal.
Height of the table \(h = \frac{1}{2}g{t^2} = \frac{1}{2} \times 10 \times {(0.4)^2} = 0.8\,m\)
Horizontal distance travelled by the ball from the edge of table \(h = ut = 4 \times 0.4 = 1.6\,m\)