3 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 \(6 \mathrm {~ms} ^ { - 1 }\) from a point 0.75 m vertically above \(O\) (see Fig. 3). When \(O A\) makes an angle \(\theta\) with the upward vertical the speed of \(A\) is \(v \mathrm {~ms} ^ { - 1 }\).
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\includegraphics[alt={},max width=\textwidth]{831ba5da-df19-43bb-b163-02bbddb4e2b8-2_388_220_1790_244}
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\caption{Fig. 3}
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- Show that \(v ^ { 2 } = 50.7 - 14.7 \cos \theta\).
- 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.