(b)

$\theta=45^{\circ}, t=1 \,s$

$\tan \phi=\frac{V_y}{U_y}=\frac{u \sin \theta-g t}{u \cos \theta}$

$\tan 45^{\circ}=\frac{u \sin \theta-g \times 1}{u \cos \theta} \Rightarrow u \cos \theta=u \sin \theta-g$

also, $V_y=0$, after $1^{\text {st }}$ (as speed is minimum)

$u \sin \theta-g \times 2=0$

$\Rightarrow u \sin \theta=2 g \ldots (i)$

so, $u \cos \theta=2 g-g$

$u \cos \theta=g \ldots (ii)$

so, $\frac{(i)}{(ii)}=\frac{u \sin \theta}{u \cos \theta}=\frac{2 g}{g}$

$\Rightarrow \tan \theta=2$

$\theta=\tan ^{-1}(2)$