6 A helicopter rescue activity at sea is modelled as follows. The helicopter is stationary and a man is suspended from it by means of a vertical, light, inextensible wire that may be raised or lowered, as shown in Fig. 6.1.

1. When the man is descending with an acceleration \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) downwards, how much time does it take for his speed to increase from \(0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards to \(3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards? \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{5211a643-307a-4886-a2e2-c11b28e05216-4_373_460_365_1242} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
   \end{figure} How far does he descend in this time? The man has a mass of 80 kg . All resistances to motion may be neglected.
2. Calculate the tension in the wire when the man is being lowered
   (A) with an acceleration of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) downwards,
   (B) with an acceleration of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) upwards. Subsequently, the man is raised and this situation is modelled with a constant resistance of 116 N to his upward motion.
3. For safety reasons, the tension in the wire should not exceed 2500 N . What is the maximum acceleration allowed when the man is being raised? At another stage of the rescue, the man has equipment of mass 10 kg at the bottom of a vertical rope which is hanging from his waist, as shown in Fig. 6.2. The man and his equipment are being raised; the rope is light and inextensible and the tension in it is 80 N .
4. Assuming that the resistance to the upward motion of the man is still 116 N and that there is negligible resistance to the motion of the equipment, calculate the \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{5211a643-307a-4886-a2e2-c11b28e05216-4_442_460_1589_1242} \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{figure} tension in the wire.