Answer
$(a)\space \sqrt {2gh+\frac{16gh}{(\sqrt 2 +4)^{2}}}$
$(b)\space 62.4^{\circ} $
Work Step by Step
Please see the image first.
(a) To the AC motion, Let's apply the equation $V^{2}=u^{2}+2aS$
$\uparrow V^{2}=u^{2}+2aS$
Let's apply known values to this equation.
$0^{2}=V^{2}+2(-g)h$
$V=\sqrt {2gh}$
To the AB motion, Let's apply equation $S=ut+\frac{1}{2}at^{2}$
$\uparrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$-h=\sqrt {2gh}\space t+\frac{1}{2}(-g)t^{2}$
$gt^{2}-2\sqrt {2gh}\space t -2h=0$
This is a quadratic formula. So we can write the results as follows.
$t=\frac{-(-2\sqrt {2gh})\space \pm \sqrt {8gh\space+8gh}}{2g}= \frac{2\sqrt {2gh} \space \pm 4\sqrt {gh}}{2g}$
Since time is a positive value, we neglect the negative value and we get,
$t=\sqrt \frac{h}{g}\space \frac{(\sqrt 2+4)}{2}$
To AB motion let's apply equation $S=ut$
$\rightarrow S=ut$
Let's plug known values into this equation.
$2h=u\sqrt \frac{h}{g}\space \frac{(\sqrt 2+4)}{2}$
$u=\frac{4\sqrt {gh}}{\sqrt 2\space +4}$
Magnitude of the initial velocity$=\sqrt {V^{2}+u^{2}}$
$\space =\sqrt {2gh+\frac{16gh}{(\sqrt 2 +4)^{2}}}$
(b) $tan\theta = \frac{V}{u}= \sqrt {2gh}\times\frac{(\sqrt 2)+4}{4\sqrt {gh}}$
$\theta = 62.4^{\circ}$
