(a) The ball has a final velocity vector
[tex]\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j[/tex]
with horizontal and vertical components, respectively,
[tex]v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}[/tex]
[tex]v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}[/tex]
The horizontal component of the ball's velocity is constant throughout its trajectory, so [tex]v_{x,i}=v_{x,f}[/tex], and the horizontal distance x that it covers after time t is
[tex]x=v_{x,i}t=v_{x,f}t[/tex]
It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:
[tex]103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s[/tex]
The vertical component of the ball's velocity at time t is
[tex]v_{y,f}=v_{y,i}-gt[/tex]
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:
[tex]-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}[/tex]
So, the initial velocity vector is
[tex]\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j[/tex]
which carries an initial speed of
[tex]\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}[/tex]
and direction θ such that
[tex]\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}[/tex]
(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is
[tex]y=v_{y,i}t-\dfrac12gt^2[/tex]
so that when it lands in the seats at t ≈ 6.38 s, it has a height of
[tex]y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}[/tex]
