Answer
$\sqrt {\frac{gL}{2\sqrt 3}}$
Work Step by Step
Let's apply equation $F=ma$ to the ball in vertical @ horizontal direction.
$\uparrow F=ma$
Let's plug known values into this equation.
$Tcos30^{\circ}-mg=m\times(0)$
$T\frac{\sqrt 3}{2}=mg=\gt T=2mg/\sqrt 3-(1)$
$\leftarrow F=ma$
Let's plug known values into this equation.
$Tsin30^{\circ}=m\times\frac{V^{2}}{r}=\gt \frac{T}{2}=m\frac{V^{2}}{r}-(2)$
$(1)=\gt(2)$
$\frac{2mg}{\sqrt 3}\times\frac{1}{2}=m\frac{V^{2}}{r}$
$\frac{g}{\sqrt 3}=\frac{V^{2}}{Lsin30^{\circ}}=\gt\frac{g}{\sqrt 3}=\frac{2V^{2}}{L}$
$V=\sqrt {\frac{gL}{2\sqrt 3}}=$ Speed of the ball
