Answer
$x_f=12.8m$
$\theta=-30.9^o$
Work Step by Step
$\Delta y=v_0 t+\dfrac{1}{2}at^2$
$0.65m=(12\frac{m}{s}\times \sin(35^o))t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$
$4.9\frac{m}{s^2}t^2-6.88\frac{m}{s}t+0.65m=0$
$t=0.102s$
$t=1.30s$
The ball reaches the height of 3.05m shortly after the shooter shoots the ball, so we need the larger time value.
$x_f=x_i+v_{xi}t+\frac{1}{2}a_xt^2$
$x_f=0+12\frac{m}{s}(1.30s)+\frac{1}{2}(0)t^2$
$x_f=12.8m$
$\theta=\arctan\Big(\frac{v_{yi}-gt}{v_{xi}}\Big)=\arctan\Big(\frac{12\times \sin(35^o)-(9.8\frac{m}{s^2})(1.30s)}{12\times \cos(35^o)}\Big)=-30.9^o$
