At what speed should the basketball be thrown?

Question

I've followed the answers of similar questions, but still seem to come up w/ the wrong answer, please help! :(

A basketball player 1.99m tall wants to make a basket from a distance 12.7m. The hoop is at a height of 3.05m.

If he shoots the ball (from a height of 1.99m) at a 61.7deg angle, at what inital speed must he throw the basketball so that it goes through the hoop w/o striking the backboard? .. answer in units m/s

Answer 1

Set up the coordinate system, such that the person held ball at (0,0) and thus the basket is at (12.7, 1.06), with meter as the unit.
Let the ball's initial speed to be V, for it to go through the hoop w/o striking the backboard. If he shoots the ball at a 61.7deg angle (my understanding is above horizon), the ball's horizontal and vertical speeds are V*cos(61.7degree) and V*sin(61.7degree), respectively.
Obviously, the time it took must be:12.7m/ [V*cos(61.7degree)] (sec)
Can you solve the problem now without looking what I would write below?
So we have an equation for the vertical distance:
1.06 = {V*sin(61.7degree)}*t - 0.5gt^2
={V*sin(61.7degree)}*12.7m/ [V*cos(61.7degree)] - 0.5*9.81*{12.7m/ [V*cos(61.7degree)]}^2
= 12.7m*tan(61.7degree) - 0.5*9.81*(12.7m)^2 / [V*cos(61.7degree)]^2
Solve V from this equation and you get the answer.