As shown in the figure, a ditch is 11.7 ft wide and has approach...

The bike acts as a projectile starting the moment its front wheel reaches the edge.

At this instant, the bike is on the incline, so the horizontal distance of its back wheel from the edge is 5 \cos 15^\{}^{{}^{\circ}}.

To safely cross, the back of the bike must clear the ditch. Thus, the total required horizontal range is:

R = 11.7 + 5 \cos 15^\{}^{{}^{\circ}}

$R \approx 11.7 + 5(0.966) = 11.7 + 4.83 = 16.53 \text{ ft}$

Using the horizontal range formula $R = \dfrac{u^2 \sin 2\theta}{g}$ with $g = 32 \text{ ft/s}^2$ and \theta = 15^\{}^{{}^{\circ}}:

16.53 = \dfrac{u^2 \sin 30^\{}^{{}^{\circ}}}{32}

$16.53 = \dfrac{u^2 (0.5)}{32}$

$u^2 = \dfrac{16.53 \times 32}{0.5} = 1057.92$

$u = \sqrt{1057.92} \approx 32.5 \text{ ft/s}$