Answer
$v_A = \sqrt {2gL}$
Work Step by Step
$v_D = 0m/s$
the energy at A = the energy at D
$U_A + K_A = U_D + K_D$
$mgh_A+ \frac{1}{2}mv_A^2 = mgh_D+ \frac{1}{2}mv_D^2 $
$\frac{1}{2}mv_A^2 = mgh_D+ \frac{1}{2}mv_D^2 - mgh_A$
$\frac{1}{2}v_A^2 = gh_D+ \frac{1}{2}v_D^2 - gh_A$
$\frac{1}{2}v_A^2 = \frac{1}{2}v_D^2+gh_D - gh_A$
$\frac{1}{2}v_A^2 = \frac{1}{2}(0)^2+g(h_D - h_A)$
$\frac{1}{2}v_A^2 = 0+g(h_D - h_A)$
$v_A^2 = 2g(h_D - h_A)$
$v_A^2 = 2g(L)$
$v_A = \sqrt {2gL}$
