A rescue plane flies to help a drowning man

Similar solution as @chakravarthy b 's

As shown in the figure:

• $\psi$ is the angle of view.
• $L$ is the horizontal distance between the plane and the man in the sea.
• $h = \text{1200 m}$ is the constant height above the sea surface the plane flies

Let $t$ be the time for the rescue bag to drop a height $h$ from the plane to the sea surface. Since $h = \dfrac 12 gt^2$ , $\implies t = \sqrt{\dfrac {2h}g}$ .

For the rescue bag dropped to be as close to the man as possible, $t$ is also the time for the plane to cover the horizontal distance $L$ with speed $v=\text{430 km/h}$ . Then $L=vt$ .

And we have:

$\begin{aligned} \tan \psi & = \frac Lh = \frac {vt}h = \sqrt{\frac 2{gh}}v = \sqrt{\frac 2{9.81 \times 1200}} \times \frac {430\times 1000}{3600} \approx 1.55688246 \\ \implies \psi & \approx \boxed{57}^\circ \end{aligned}$

The initial velocity of the bag is identical to the velocity of the plane. Therefore it has the magnitude v .

If we neglect the resistance of environment, the bag moves vertically in a free fall.

Because we know at what height it was released, we can easily determine the duration of its fall to the surface. It holds in the direction of the y-axis that:

$-h\,=\,-\,\frac{1}{2}gt^{2}\,,$

therefore:

$h\,=\,\frac{1}{2}gt^{2}\,.$

(only the gravitational force acts upon the bag in the direction of the y-axis)

By solving this equation we obtain for the duration of the flight t:

$t\,=\,\sqrt{\frac{2h}{g}}\,.$

The bag (and the plane) move horizontally with constant speed v .

From the previous hint we see that for the duration of the fall to the surface t holds:

$t\,=\,\sqrt{\frac{2h}{g}}\,.$

In this time, the bag travels the horizontal distance:

$L\,=\,vt\,.$

By substituting for t we obtain:

$L\,=\,v\,\sqrt{\frac{2h}{g}}\,.$

The way to calculate the angle of view ψ is obvious from the diagram.

We will use the trigonometric function tan

In our calculation, we need to know the distance L the bag will travel horizontally during the time t before it hits the surface and the height from which it falls. We know both.

It holds that:

$\mathrm{tan}\psi\,=\,\frac{L}{h}\,=\,\frac{v\,\sqrt{\frac{2h}{g}}}{h}\,=\,v\,\sqrt{\frac{2}{hg}}\,.$

Numerically:

$\mathrm{tan}\psi\,=\frac{430\, 000}{3\, 600}\sqrt{\frac{2}{1\, 200{\cdot}9.81}}\,\dot=\, 1.556\,,$

$\psi\,=\,57^{\circ}\,.$

Note: the horizontal projection of the velocity of the bag is at every moment identical with the velocity of the plane, so the pilot always sees the falling bag directly below him.