Messi you can do it

The 2nd equation of motion in vertical direction for the ball is

$H=v\sin \theta t -\frac {1}{2} gt^{2}$

$\frac { 1 } { 2 } gt^ { 2 } -v \ sin \theta t+H=0$

$t_{1}=\frac {v\sin\theta - \sqrt {v^{2}(\sin\theta)^{2}-2gh} }{g}$ and $t_{2}=\frac {v\sin\theta +\sqrt {v^{2}(\sin\theta)^{2}-2gh} }{g}$

These are the times when the ball is at above the hands of gk and below the bar respectively. Their difference gives the time the ball is between gk and the goal.

$T=t_{2}-t_{1}=\frac {2\sqrt {v^{2}(\sin \theta)^{2}-2gh} }{g}$

As the distance between the gk and the goal is H

$v\cos \theta ×T = H$

Squaring the equation above gives

$4 (v^{2} (\sin \theta)^{2}-2gH)v^{2}(\cos \theta )^{2}=(gh)^{2}$

Putting $v=\sqrt {zgh}$ and $\theta = 45$ we get

$z^{2}-4z-1=0$

$z= 2+\sqrt {5}$

$\boxed {x=2 ,y=5}$ Ans. Is $\boxed { 29}$

What do you think, will barca win?

We will ignore air friction and the ball's dimensions, as nothing is mentioned about that. With pain in the heart we also ignore Messi's use of effect. The ball's trajectory then is a parabola that is parametrized by $\left\{ \begin{array}{ll} x(t)=x_0+v_{x0}t\\ y(t)=y_0+v_{y0}t -\frac {1}{2} gt^2\\ \end{array} \right.$

Because the ball is shot at $t=0$ from $y=0$ , at an angle of $45°$ , we know $y_0 =0 \text{ and } v_{x0}^2=v_{y0}^2=\frac{1}{2}v^2$

The ball goes from Messi's foot at $y=0$ , via Buffon's fingertips at $y=h$ to the bar of the goal at $y=h$ . The top of the parabola is therefore midway between Buffon and the goal.

Let $\Delta t$ be the time that it takes from passing Buffon's fingertip to reaching the top of the trajectory. The horizontal distance is $\frac12h$ so that $\Delta t=\frac {h}{2v_{x0}}$

Let $\Delta y$ be the height of the trajectory's top above $h$ , then $\Delta y=\frac{1}{2}g(\Delta t)^2=\frac{gh^2}{8v_{x0}^2}=\frac{gh^2}{4v^2}$ , so that $y_{top}=h+\frac{gh^2}{4v^2}$

But we can find another expression for $y_{top}$ . By setting $y'(t)=v_{y0} -gt=0$ we get $t_{top}=\frac{v_{y0}}{g}$ and $y_{top}=0+v_{y0}\frac{v_{y0}}{g} -\frac {1}{2} g(\frac{v_{y0}}{g})^2=\frac{v_{y0}^2}{2g}=\frac{v^2}{4g}$

Set both expressions for $y_{top}$ equal, and multiply everything by $v^2$ , to get a quadratic expression in $v^2$ : $\frac{1}{4g}v^4-hv^2-\frac{gh^2}{4}=0$ from which we solve $v^2=\frac{h \pm \sqrt{h^2+h^2/4}}{\frac{2}{4g}}=gh(2\pm\sqrt{5})$ so that, using the real solution, $v=\sqrt{gh(2+\sqrt{5})}$ and the asked expression is $2^2+5^2=\boxed{29}$ .

Writing the equation of trajectory of the ball

y = x - gx^2 / u^2

When y = H we will get two values of x whose difference of roots should be H .

Solving the above constraint we will get desired answer :)