2 One end of a light inextensible string of length 0.75 m is attached to a particle \(A\) of mass 2.8 kg . The other end of the string is attached to a fixed point \(O . A\) is projected horizontally with speed \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point 0.75 m vertically above \(O\) (see Fig. 2). When \(O A\) makes an angle \(\theta\) with the upward vertical the speed of \(A\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
\(\xrightarrow [ A \text { a } ] { 6 \mathrm {~m} \mathrm {~s} ^ { - 1 } }\) Fig. 2

1. Show that \(v ^ { 2 } = 50.7 - 14.7 \cos \theta\).
2. Given that the string breaks when the tension in it reaches 200 N , find the angle that \(O A\) turns through between the instant that \(A\) is projected and the instant that the string breaks.