The ball's vertical position [tex]y[/tex] in the air at time [tex]t[/tex] is
[tex]y=v_0\sin31^\circ\,t-\dfrac g2t^2[/tex]
The ball is at its original height when [tex]y=0[/tex], which happens at
[tex]v_0\sin31^\circ\,t-\dfrac g2t^2=\dfrac t2\left(2v_0\sin31^\circ-gt)=0[/tex]
[tex]\implies t=0\text{ and }t=\dfrac{2v_0\sin31^\circ}g[/tex]
Meanwhile, the ball's horizontal position [tex]x[/tex] at time [tex]t[/tex] is
[tex]x=v_0\cos31^circ\,t[/tex]
So when the ball reaches its original height a second time, the ball will have traveled a horizontal distance of
[tex]x=\dfrac{2{v_0}^2\sin31^\circ\cos31^\circ}g=\dfrac{{v_0}^2\sin(2\cdot31^\circ)}g[/tex]
(which you might recognize as the formula for the range of a projectile)
To reach a distance of [tex]x=77\,\rm m[/tex], the initial speed [tex]v_0[/tex] would be
[tex]77\,\mathrm m=\dfrac{{v_0}^2\sin62^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\implies v_0=29\dfrac{\rm m}{\rm s}[/tex]
