College Physics
posted by Taylor on .
A 0.35 kg stone attached to a 0.8 m long string is rotated in a horizontal plane. The string makes an angle of 20° with the horizontal. Determine the speed of the stone.
I've tried tan 20= gr/v^2 = (9.8)(.8)/v^2 = 4.7 m/s and that's not the correct answer (online homework). Any help or explanation would be appreciated

mg=Tsin α
mv²/R=Tcos α
mgR/ mv² = Tsin α/Tcos α
gR/ v² =tan α
gLcos α/ v² =tan α
v=sqrt{ gLcos α/ tan α} =sqrt{9.8•0.8•cos20°/tan20°} = 4.5 m/s