Speed at lowest point, v1=6m∕s Angle made by string with vertical, θ=60° In Δ‌OBC,
OC
OB
=cos‌θ=cos‌60°
l−h
l
=cos‌60° ⇒h=l(1−cos‌60°) ⇒h=0.5(1−
1
2
)=0.25m Using law of conservation of mechanical energy, we get Ki+Ui=Kf+Uf ⇒
1
2
mv12+0=
1
2
mv22+mgh [Taking initial potential energy at ground is equal to zero.] ⇒
1
2
mv22=
1
2
mv12−mgh ⇒v22=v12−2gh Substituting the values, we get v2=√(6)2−2×10×0.25=√31m∕s Hence, the final velocity at an angle 60° is √31m∕s.