If the ball is released with velocity v at (x, y) = (0, 1.97) and must pass through the point (11.6, 3.05), the horizontal and vertical equations of motion can be written as

x = v*because(40.2°)*t

y = 1.97 + v*sin(40.2°)*t - (1/2)*9.8*t^2

where the acceleration due to gravity is taken to be -9.8 m/s^2.

We can solve the first equation for t and substitute that expression into the second equation. At the hoop position, we have

11.6 = v*.763796*t

15.1873/v = t (divide by the coefficient of t)

3.05 = 1.97 + v*.645458*(15.1873/v) - 4.9*(15.1873/v)^2

3.05 = 1.97 + 9.80276 - 1130.205/v^2 (simplify)

-8.72276 = -1130.205/v^2 (subtract the constant terms)

v^2 = 1130.205/8.72276 (multiply by v^2/-8.72276)

v = √129.570 (take the square root)

The ball is in the air about 15.1873/11.3829 = 1.334 seconds, computed using our expression for t.

x = v*because(40.2°)*t

y = 1.97 + v*sin(40.2°)*t - (1/2)*9.8*t^2

where the acceleration due to gravity is taken to be -9.8 m/s^2.

We can solve the first equation for t and substitute that expression into the second equation. At the hoop position, we have

11.6 = v*.763796*t

15.1873/v = t (divide by the coefficient of t)

3.05 = 1.97 + v*.645458*(15.1873/v) - 4.9*(15.1873/v)^2

3.05 = 1.97 + 9.80276 - 1130.205/v^2 (simplify)

-8.72276 = -1130.205/v^2 (subtract the constant terms)

v^2 = 1130.205/8.72276 (multiply by v^2/-8.72276)

v = √129.570 (take the square root)

**v = 11.38 m/s**_____The ball is in the air about 15.1873/11.3829 = 1.334 seconds, computed using our expression for t.