An unknown pitcher

$X$ -axis

$V_0 \cos\theta t=2x$

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$Y$ -axis

$0=V_{0}-gt_1$

$V_{0}=gt_1$

$0^2=(gt_1)^2-2g(x)$

$(gt)^2=2g(x)$

$gt_1^{2}=2x$

$t_1=\sqrt{\frac{2x}{g}}$ (time: he throws-touch the roof)

${V_f}^2=-2g(-2x)$ (since the roof until floor)

${V_f}=0-gt_2$

$4xg=g^{2}t_2^{2}$

$t_2=\sqrt{\frac{4x}{g}}$

Total time

$t_t=\sqrt{\frac{4x}{g}}+\sqrt{\frac{2x}{g}}$

$V_0\sin \theta=gt_1$

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$\frac{V_0 \sin \theta}{V_0 \cos\theta t_t}=\frac{gt_1}{2x}$

$\frac{V_0 \sin \theta}{V_0 \cos\theta }=\frac{gt_1t_t}{2x}$

$\frac{V_0 sin \theta}{V_0 cos\theta }=1+\sqrt{2}$

$\tan\theta=1+\sqrt{2}$

$\theta= 67.5°$

Let the initial velocity and the angle of throw with the horizon be $v$ and $\theta$ respectively.

If the ball touches the ceiling at time $t = t_1$ , then:

$0 = v\sin{\theta} - gt_1 \quad \Rightarrow v\sin{\theta} = gt_1$

Now, the vertical displacement $y(t)$ at time $t$ is given by:

$y(t) = v\sin{\theta}t - \frac{1}{2}gt^2 = v\sin{\theta}\left( t - \dfrac{t^2}{2t_1} \right)$

$\Rightarrow y(t_1) = x = v\sin{\theta}\left( t_1- \dfrac{t_1^2}{2t_1} \right) = \frac {1}{2} v\sin{\theta} t_1$

Let the ball hit point $B$ at $t_2$ , then:

$\Rightarrow y(t_2) = -x = v\sin{\theta}\left( t_2- \dfrac{t_2^2}{2t_1} \right)$ and since $x = \frac {1}{2} v\sin{\theta} t_1$ .

$\Rightarrow v\sin{\theta}\left( t_2- \dfrac{t_2^2}{2t_1} \right) = - \frac {1}{2} v\sin{\theta} t_1 \quad \Rightarrow v\sin{\theta}\left( t_2- \dfrac{t_2^2}{2t_1} + \dfrac {t_1}{2} \right) = 0$

$\Rightarrow t_2- \dfrac{t_2^2}{2t_1} + \dfrac {t_1}{2} = 0 \quad \Rightarrow t_2^2 - 2t_1t_2 - t_1^2 = 0$

$\Rightarrow t_2 = \dfrac {2t_1 + \sqrt{4t_1^2+4t_1^2}}{2} = t_1(1+\sqrt{2})$

We know that: $v\cos{\theta}t_2 = 2x \quad \Rightarrow x = \frac {1}{2} v \cos {\theta} t_1(1+\sqrt{2})$ .

$\Rightarrow x = \frac {1}{2} v\sin{\theta} t_1 = \frac {1}{2} v \cos {\theta} t_1(1+\sqrt{2})\quad \Rightarrow \tan {\theta} = 1+\sqrt{2}$

$\Rightarrow \theta = \tan^{-1} {1+\sqrt{2}} = \boxed{67.5^\circ}$

Let me give every one a better graphic.