એક ટેબલ પરથી એક પદાર્થને 4 m/sec ના સમક્ષ

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 $h = ut = 4 \times 0.4 = 1.6\,m$