We would find the time at which the ball arrives the goal from the constant velocity model in the x direction. We have
xf = xi + vit
vit = xf - xi
t = (xf - xi)/vi
where
xf is the final position of ball
xi is the initial position of ball
vi is the initial velocity of the ball in the x direction
t is the time
From the information given,
xi = 0
xf = 36
vxi = vicosθ
θ = 47
v = 20.6
vxi = 20.6cos47
t = (36 - 0)/20.6cos47
t = 2.56 s
We would find the height of the ball at time of t = 2.56 from the particle under constant acceleration model in the y direction. We have
yf = yi + vyit + 1/2ayt^2
From the information given,
yf = 0 + 20.6sin47 * 2.56 + 1/2 * 9.81 * 2.56^2
yf = 38.569 - 32.145
yf = 6.42
The ball clears the crossbar by 6.42 - 3.05
= 3.27 m