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Let u be the speed at which the basketball leaves the hand of the player.
Horizontally: 9.144 = [u.cos(46)].t
 where t is the time of flight of the basket ball.
hence, t = 9.144/(u.cos(46))   ----------------- (1)

Vertically: (3.05 - 1.9) = (u.sin46).t + (1/2)gt^2
i.e. 1.15 = (u.sin(46).[9.144/(u.cos(46)] + (-10/2).[9.144/(u.cos(46)]^2
[take g, the acceleration due to gravity, to be 10 m/s^2]

1.15 = 9.144.tan(46) - 5 x 9.144^2/(u.cos(46))^2
solve for u^2 gives u^2 = 104.1 (m/s)^2

Kinetic energy (translational) of ball on leaving the player's hand
= (1/2).(0.65).(104.1) J = 33.83 J

A NBA size basket ball has a radius of about 12 cm
Moment of inertia of basket ball = (2/3).(0.65).(0.12)^2 kg.m^2 = 0.00624 kg.m^2
Hence, rotational kinetic energy of ball
= (1/2).(0.00624).(4.pi)^2 J = 0.4927 J

Total kinetic energy of ball on leaving the player's hand
= (33.83 + 0.4927) J = 34.32 J

Assume on throwing the ball, the player pushes his hand out a distance of 0.5 m.
Thus,  F.(0.5) = 34.32   where F is the pushing force given to the ball
F = 68.65 N

Therefore, the player needs to exert a pushing force of 69 N to throw the ball.

Please be aware that the pushing force is in additional to the weight of the ball, which the player needs to exert in holding the ball. Also, air resistance has been neglected in the calculation. The actual pushing force required should be higher.