I assume you mean an angle of θ, and not 0. I also assume the given components of the initial velocity are horizontal and vertical, respectively, so that
1. the initial velocity vector is
[tex]\boxed{\vec v_i = (7.5\,\vec\imath + 13\,\vec\jmath) \dfrac{\rm m}{\rm s}}[/tex]
which means the ball is thrown with an initial speed of
√((7.5 m/s)² + (13 m/s)²) ≈ 15 m/s,
and
2. the angle made by [tex]\vec v_i[/tex] with the positive horizontal axis is θ such that
[tex]\tan(\theta) = \dfrac{13}{7.5} \implies \theta \approx \arctan(1.7) \approx \boxed{60^\circ}[/tex]
3. At time t, the ball attains a height y and horizontal range x according to
y = (13 m/s) t - g/2 t²
x = (7.5 m/s) t
where g = 9.8 m/s². When the ball reaches the ground (y = 0) for t > 0, we have
(13 m/s) t - g/2 t² = 0
13 m/s - g/2 t = 0
t = 2 (13 m/s)/g = (26 m/s)/g
Plugging this time into the x equation gives a horizontal range of
x = (7.5 m/s) (26 m/s)/g ≈ 20. m
