In Fig. 33-77, an albatross glides at aconstant $\mathbf{15}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$horizontallyabove level ground, moving in a vertical plane that contains the Sun. It glides toward a wallof height$\mathit{h}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathit{m}$, which it will just barely clear. At that time of day, the angle of the Sun relative to the groundis$\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{°}$.At what speed does the shadow of the albatross move (a) across the level ground and then (b) up the wall? Suppose that later a hawk happens to glide along the same path, alsoat$\mathbf{15}\mathit{m}\mathbf{/}\mathit{s}$.You see that when its shadow reaches the wall, the speed of the shadow noticeably increases. (c) Is the Sun now higher or lower in the sky than when the albatross flew by earlier? (d) If the speed of the hawk’s shadow on the wallis$\mathbf{45}\mathit{m}\mathbf{/}\mathit{s}$,what is the angle u of the Sun just then?

Speed of albatross is$15m/s$

$h=2.0m$

$\theta =30°$

Speed of hawk is$15m/s$

Speed of hawk’s shadow on the wall is$45m/s$

As the sun is far away, we consider that the rays coming from the sun are parallel, i.e., the angle between the sunray and the bird’s position is the same everywhere.

Formula:

$\text{tan\hspace{0.17em}}\theta =y/x$

$v=\frac{dx}{dt}$

The rays coming from the sun are parallel. So, the angle formed by the sunrays with the bird at one position is the same as the one formed with the bird at another position.

Therefore, the shadow of the bird moves on the ground with the same speed as that of the bird.

Thus, the shadow of the albatross moves across the level ground with speed$15m/s$

Now the bird is in the position$x>0$from the wall. So, its shadow on the wall is at a distance$0\ge y\ge h$from the top of the wall.

From the figure,

$\begin{array}{c}\text{tan\hspace{0.17em}}\theta =\frac{y}{x}\\ y=x\text{tan}\theta \end{array}$

Thus,

$\begin{array}{c}\frac{dy}{dt}=\frac{dx}{dt}\text{tan\hspace{0.17em}}\theta \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)\\ \frac{dy}{dt}=(-15)\text{tan\hspace{0.17em}}30°\\ =(-15)\times 0.577\\ =-8.7m/s\end{array}$

This means that the distanceyi.e., the positive number downward from the top of the wall is shrinking at the rate of 8.7 m/s.

Thus, the speed at which the shadow of the albatross moves up the wall is.$8.7m/s$

$\text{tan\hspace{0.17em}}\theta$is directly proportional to the value of the $\theta$in the range.$0°<\theta <90°$

From equation (1), it is clear that the value of $\frac{dy}{dt}$is maximum when$\text{tan\hspace{0.17em}}\theta$ is maximum. Thus, the larger value of$\left|\frac{dy}{dt}\right|$ implies a larger value of.$\theta$

Thus, we can conclude that the sun is higher in the sky when the hawk glides by.

$\frac{dy}{dt}=45m/s$

From equation (1), we have

$\begin{array}{c}\frac{dy}{dt}=\frac{dx}{dt}\text{tan\hspace{0.17em}}\theta \\ 45=\frac{dx}{dt}\text{tan\hspace{0.17em}}\theta \end{array}$

The speed of hawk is $\frac{dx}{dt}=15m/s$

Therefore,

$\begin{array}{c}45=15\text{tan}\theta \\ \text{tan}\theta =\frac{45}{15}\\ \theta =\left(71.56\right)°\\ \approx \left(72\right)°\end{array}$

The angleof the Sun if the speed of the hawk’s shadow on the wall is 45 m/s is$\theta =72°$