Let v denote the initial speed of the ball. The ball's position at time t is given by the vector
[tex]\mathbf r(t)=v\cos40^\circ\,t\,\mathbf i+\left(v\sin40^\circ\,t-\dfrac g2t^2\right)\,\mathbf j[/tex]
where g is the acceleration due to gravity with magnitude 9.80 m/s^2.
The ball reaches the goal 13 m away at time t such that
[tex]10\,\mathrm m=v\cos40^\circ t\implies t=\dfrac{10\,\mathrm m}{v\cos40^\circ}[/tex]
at which point it attains a height of 2.44 m, so that
[tex]2.44\,\mathrm m=v\sin40^\circ\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)-\dfrac g2\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)^2[/tex]
[tex]2.44\,\mathrm m=(10\,\mathrm m)\tan40^\circ-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)\left(\dfrac{100\,\mathrm m^2}{v^2\cos^240^\circ}\right)[/tex]
[tex]\implies\boxed{v\approx3.75\dfrac{\rm m}{\rm s}}[/tex]
