Ball kicked, find launch speed

Treat the ball as a projectile. Use ft units with $g=32.2\ \text{ft/s}^2$.

Equations of motion (origin at A)

$$x=(v_A\cos30^\circ)t,\qquad y=(v_A\sin30^\circ)t-\tfrac12 g t^{2}$$

At B: $x=15$ ft, $y=-9$ ft. Eliminating $t$:

$$y=x\tan\theta-\frac{g x^{2}}{2v_A^{2}\cos^{2}\theta}$$ $$-9=15\tan30^\circ-\frac{32.2(15)^{2}}{2v_A^{2}\cos^{2}30^\circ}$$

Since $32.2(15)^2=7245$ and $2\cos^2 30^\circ=1.5$:

$$-9=8.660-\frac{7245}{1.5\,v_A^{2}}=8.660-\frac{4830}{v_A^{2}}\ \Rightarrow\ \frac{4830}{v_A^{2}}=17.660$$ $$v_A^{2}=273.5\ \Rightarrow\ v_A=16.54\ \text{ft/s}$$

Check: $t=\dfrac{15}{16.54\cos30^\circ}=1.047$ s.

Answer: the ball is kicked at $v_A\approx16.5$ ft/s (time of flight $\approx1.05$ s).