Amount of water inside the horizontal cylindrical tank

$\cos \frac{1}{2}\theta = \dfrac{0.5}{2.5}$

$\frac{1}{2}\theta = 78.463^\circ$

$\theta = 156.926^\circ$
 

$\alpha = 360^\circ - \theta = 360^\circ - 156.926^\circ$

$\alpha = 203.074^\circ$
 

Area of cross section:
$A_b = \frac{1}{2} r^2(\alpha_{rad} + \sin \theta_{deg})$

$A_b = \frac{1}{2} (2.5^2)\left( 203.074^\circ \cdot \dfrac{\pi}{180^\circ} + \sin 156.926^\circ \right)$

$A_b = 12.3 ~ \text{ft}^2$
 

Volume of water:
$V_w = A_b L = 12.3(12)$

$V_w = 147.61 ~ \text{ft}^3$
 

Depth of water in vertical position:
 

$V_w = \pi r^2 h$

$147.61 = \pi (2.5^2)h$

$h = 7.518 ~ \text{ft}$           answer