To determine the total distance covered by the bird, let's proceed step-by-step.

First, we need to calculate the time it takes for the two cars to meet. The two cars are moving towards each other, so their relative speed is the sum of their individual speeds.

The speed of the first car is 18 km/h and the speed of the second car is 27 km/h. Therefore, their relative speed is:

$$18 \, \text{km/h} + 27 \, \text{km/h} = 45 \, \text{km/h}$$

The initial distance between the two cars is 36 km. To find the time it takes for the cars to meet, we use the formula:

$$\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}}$$

Substituting the provided values, we get:

$$\text{Time} = \frac{36 \, \text{km}}{45 \, \text{km/h}} = \frac{4}{5} \, \text{hours} = 0.8 \, \text{hours}$$

Now, we know that the bird is flying at a speed of 36 km/h. To find the total distance covered by the bird while the two cars meet, we use the formula:

$$\text{Distance} = \text{Speed} \times \text{Time}$$

Substituting the speed of the bird and the time calculated, we get:

$$\text{Distance} = 36 \, \text{km/h} \times 0.8 \, \text{hours} = 28.8 \, \text{km}$$

Thus, the total distance covered by the bird is 28.8 kilometers.

Given, horizontal speed of the man ($u_x$) = 9 m/s

Horizontal distance between the two buildings = 10 m

Height difference between the two buildings = 9 m

and $g = 10 \text{ m/s}^2$

Let the man jumps from point $A$ and land on the roof of the next building at point $B$.

Taking motion in vertical direction,

$y = ut + \frac{1}{2} at^2$

$9 = 0 \times t + \frac{1}{2} \times 10 \times t^2$

$9 = 5t^2$

or $t = \sqrt{ \frac{9}{5}} = \frac{3}{\sqrt{5}}$

Therefore, horizontal distance travelled = $u_x \times t = 9 \times \frac{3}{\sqrt{5}} = \frac{27}{\sqrt{5}} \approx 12$ m

Horizontal distance travelled by the man is greater than 10 m, therefore, he will land on the next building.

For the ball dropped from the building, $u_1 = 0, u_2 = 40 \text{ m/s}$
Velocity of the dropped ball after time $t$,

$v_1 = u_1 + gt$

$v_1 = gt$ (downward)

For the ball thrown up, $u_2 = 40 \text{ m/s}$

Velocity of the ball after time $t$:

$v_2 = u_2 - gt$

$v_2 = (40 - gt) $ (upward)

∴ Relative velocity of one ball with respect to another ball:

$v_1 - v_2$

$= gt - [40 - gt]$

$= 40 \text{ m/s}$

Note: When we are applying equations for rectilinear motion we should carefully put up the signs for the physical quantities.

Given, initial velocity $= v_0$

Let the distance traveled in time $t = x_0$.

For the graph:

$ \tan \theta = \frac{v_0}{x_0} = \frac{v_0 - v}{x}$ ...($i$)

where, $v$ is velocity, and $x$ is the displacement at any instant of time $t$.

From Eq. ($i$):

$v_0 - v = \frac{v_0}{x_0} x$

⇒ $v = -\frac{v_0}{x_0} x + v_0$

We know that

$$ \begin{aligned} \text { Acceleration } a & =\frac{d v}{d t}=\frac{-v_0}{x_0} \frac{d x}{d t}+0 \\\\ \Rightarrow a & =\frac{-v_0}{x_0}(v) \\\\ & =\frac{-v_0}{x_0}\left(\frac{-v_0}{x_0} x+v_0\right) \\\\ & =\frac{v_0{ }^2}{x_0{ }^2} x-\frac{v_0{ }^2}{x_0} \end{aligned} $$

Graph of a versus $x$ is given above.

In this problem equation of motion and Newton’s seconds law that is $F_{ext} = \frac{dp}{dt}$ will be used, where dp is change in momentum over time dt.

Given, height $(h) = 1 km = 1000 m$

$g = 10 m/s^2$

(a) Velocity attained by the rain drop in freely falling through a height $h$.

$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1000} = 100 \sqrt{2}$ m/s

$= 100 \sqrt{2} \times \frac{60 \times 60}{1000}$ km/h

$= 360 \sqrt{2}$ km/h $\approx 510$ km/h

(b) Diameter of the drop $(d) = 2r = 4 mm$

$\therefore$ Radius of the drop $(r) = 2$ mm $= 2 \times 10^{-3}$ m

Mass of a rain $\operatorname{drop}(m)=V \times \rho$

$$ \begin{aligned} & =\frac{4}{3} \pi r^3 \rho \\ & =\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3 \quad\left(\because \text { Density of water }=10^3 \mathrm{~kg} / \mathrm{m}^3\right) \\ & \approx 3.4 \times 10^{-5} \mathrm{~kg} \end{aligned} $$

Momentum of the rain $\operatorname{drop}(p)=m v$

$$ \begin{aligned} & =3.4 \times 10^{-5} \times 100 \sqrt{2} \\ & =4.7 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \\ & \approx 5 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \end{aligned} $$

(c) Time required to flatten the drop = time taken by the drop to travel the distance equal to the diameter of the drop near the ground

$$ \begin{aligned} t & =\frac{d}{V}=\frac{4 \times 10^{-3}}{100 \sqrt{2}}=0.028 \times 10^{-3} \mathrm{~s} \\ & =2.8 \times 10^{-5} \mathrm{~s} \approx 30 \mathrm{~ms} \end{aligned} $$

(d) Force exerted by a rain drop

$$ \begin{aligned} F & =\frac{\text { Change in momentum }}{\text { Time }}=\frac{p-0}{t} \\ & =\frac{4.7 \times 10^{-3}}{2.8 \times 10^{-5}} \approx 168 \mathrm{~N} \end{aligned} $$

(e) Radius of the umbrella $(R)=\frac{1}{2} \mathrm{~m}$

$\therefore$ Area of the umbrella $(A)=\pi R^2=\frac{22}{7} \times\left(\frac{1}{2}\right)^2=\frac{22}{28}=\frac{11}{14} \approx 0.8 \mathrm{~m}^2$

Number of drops striking the umbrella simultaneously with average separation of 5 cm $=5 \times 10^{-2} \mathrm{~m}$.

$$ =\frac{0.8}{\left(5 \times 10^{-2}\right)^2}=320 $$

$\therefore$ Net force exerted on umbrella $=320 \times 168=53760 \mathrm{~N} \approx 54000 \mathrm{~N}$

Note In practice, the velocity of the drops decreases due to air friction.