Answer
The minimum angle is
$$\alpha_0=15.5^{\circ}.$$
The initial speed is
$$v_0=12.2 \text{ m/s}=43.9 \text{ km/s}.$$
Work Step by Step
Firstly, let's convert English units to SI units.
$36.0 \text{ ft} = 10.97 \text{ m}$, $10.0 \text{ ft} = 3.048 \text{ m}$.
(a) No matter how the ball is kicked it can't be aimed lower than directly at the bar. So the minimum angel is
$\tan\alpha_0=\dfrac{3.048}{36.0}$ or $\alpha_0=15.5^{\circ}$.
(b) Let's choose a coordinate system as shown in the figure. Initial conditions are
y-component:
$y_0=0,$
$v_{0y}=v_0\sin45^{\circ}$
$a_y=-9.8 \text{ m/}\text{s}^2$
x-component:
$x_0 = 0,$
$v_{x0}=v_0\cos45^{\circ}$
$a_x=0$
So we have
y-component:
$y=v_0\sin45^{\circ}t - \dfrac{9.8t^2}{2}$
$v_y=v_0\sin45^{\circ} - 9.8t$
x-component:
$x=v_0\cos45^{\circ}t$
$v_x=v_0\cos45^{\circ}$
At the time the ball clear the bar its x-coordinate is $L=10.97 \text{ m}$. Let's find this time.
$t=\dfrac{L}{v_0\cos45^{\circ}}$
The ball's y-coordinate at the time is $H=3.048 \text{ m}$
$3.048=v_0\sin45^{\circ}\cdot\dfrac{10.97}{v_0\cos45^{\circ}}-\dfrac{9.8}{2}\left(\dfrac{10.97}{v_0\cos45^{\circ}}\right)^2$
or
$v_0=\dfrac{10.97}{\cos45^{\circ}}\left(\dfrac{9.8}{2\cdot10.97\tan45^{\circ}-2\cdot3.048}\right)^{1/2}$
$v_0=12.2 \text{ m/s}$ or $v_0=43.9 \text{ km/h}$.
