I need help with the physics question that I can't get correctly. (Pendulum)

Question

idku · Answer

A pendulum is constructed by attaching a small metal ball to one end of a L=1.60-m-long string that hangs from the ceiling (see the figure below). The ball is released when it is raised high enough for the string to make an angle of 30.0° with the vertical.

How fast is it moving at the bottom of its swing?

idku · Answer

Well, I wanted to apply
$v=\sqrt{2gh}$

So, I got $v=\sqrt{2*9.80(m/s^2)*1.60(m)}=\sqrt{31.36~(m^2/s^2)}=5.6~m/s$

idku · Answer

but the site tells me that I am not correct.

malijhaa · Answer

Um So Do Yhu Still Need Help?

pucusana · Answer

if i help you would you help me?

idku · Answer

yes, I do need help

pucusana · Answer

i think you might be correct 5.6 m/s try and double check

pucusana · Answer

i have a question

pucusana · Answer

try and find my question

irishboy123 · Answer

little tweak!

$v=\sqrt{2*9.80(m/s^2)*1.60(m) \color{red}{(1 - \cos(30^o))}}= \dots$