(a) The ball has a final velocity vector

with horizontal and vertical components, respectively,


The horizontal component of the ball's velocity is constant throughout its trajectory, so
, and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

The vertical component of the ball's velocity at time <em>t</em> is

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

So, the initial velocity vector is

which carries an initial speed of

and direction <em>θ</em> such that

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of
