Question

A child in danger of drowning in a river is being carried downstream by a current that flows uniformly at a speed of $\frac{16}{3} m s^{- 1}$ . The child is $600 m$ from shore and $800 m$ downstream of a helipad when the rescue helicopter sets out and the helicopter follows the same direction throughout the motion. The helicopter proceeds at its maximum speed of $\frac{80}{3} m s^{- 1}$ with respect to air while air is blowing with a velocity of $4 m s^{- 1}$ perpendicular to the river with water flowing velocity as shown. Then,
(i) heading at what angle with the shore should the helicopter take off?
(ii) calculate the time taken by the helicopter to reach the child.

Accepted Answer

(i) Given, the velocity of river flow, $(u) = \frac{16}{3} m s^{- 1}$ ,
velocity of helicopter with respect to air, $(u_{H}) = \frac{80}{3} m s^{- 1}$ ,
velocity of air, $(v_{a}) = 4 m s^{- 1}$ .

Suppose the time taken by helicopter to reach the boy is $t$ .

Here, we can infer from the diagram that the final position of both the child and the helicopter must be the same (at $C$ ). Suppose the helicopter is moving at an angle $α$ from the flow of river.
So, the distance travelled by the helicopter in the direction of flow will be equal to the sum of distance travelled by the boy and the initial separation between the helicopter and the boy in the direction of flow.

$∴ (u_{H} \cos (α)) t = u t + 800$ ,

$\Rightarrow (\frac{80}{3} \cos (α)) t = \frac{16}{3} t + 800 \dots (i)$ .

The distance travelled by the helicopter with respect to the ground in the direction perpendicular to the flow will be equal to the initial separation between the helicopter and the boy in the direction perpendicular to the flow.

$∴ (u_{H} \sin (α) - v_{a}) t = 600$ ,

$\Rightarrow (\frac{80}{3} \sin (α) - 4) t = 600 \dots (ii)$ .

Solving equations $(i) and (ii)$ , the angle at which helicopter is flying, $(α) = 37 °$ .

(ii) Put the value of angle $α$ in equation $(ii)$ , we get,

$(\frac{80}{3} \sin (37 °) - 4) t = 600$ .

Time taken, $(t) = 50 s$ .

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