A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?
Consider the adjacent diagram. Let a fighter plane, when it be at position $ P $, drops a bomb to hit a target $ T $.
Let
$ \angle P'\,PT = \theta $
Speed of the plane $ = 720\, \text{km/h} = 720 \times \frac{5}{18} \text{ m/s} = 200 \text{ m/s} $
Altitude of the plane $(P'T) = 1.5\, \text{km} = 1500\, \text{m} $
If bomb hits the target after time $ t $, then horizontal distance travelled by the bomb,
$ PP' = u \times t = 200t \hspace{30mm} \text{...(i)}$
Vertical distance travelled by the bomb,
$ P'T = \frac{1}{2} gt^2 \implies 1500 = \frac{1}{2} \times 9.8t^2 $
$ \implies t^2 = \frac{1500}{4.9} \implies t = \sqrt{\frac{1500}{49}} = 17.49\, \text{s} $
Using value of $ t $ in Eq. (i),
$ PP' = 200 \times 17.49\, \text{m} $
Now,
$ \tan \theta = \frac{P'T}{P'P} = \frac{1500}{200 \times 17.49} = 0.49287 = \tan 23^\circ 12' $
$ \theta = 23^\circ 12' $
Note Angle is with respect to target. As seen by observer in the plane motion of the bomb will be vertically downward below the plane.
(a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of the earth due to the earth rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre? What is it at latitude 0°? How does these accelerations compare with $g = 9.8 \text{ m/s}^2$?
(b) Earth also moves in circular orbit around the sun once every year with an orbital radius of $1.5 \times 10^{11}$ m. What is the acceleration of the earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with $g = 9.8 \text{ m/s}^2$?
$\begin{array}{ll} \text{(a)} & \text{Radius of the earth } (R) = 6400 \text{ km} = 6.4 \times 10^6 \text{ m} \\ & \text{Time period } (T) = 1 \text{ day} = 24 \times 60 \times 60 \text{ s} = 86400 \text{ s} \\ & \\ & \text{Centripetal acceleration } (a_c) = \omega^2 R = R \left(\frac{2\pi}{T}\right)^2 = \frac{4\, \pi^2\, R}{T^2} \\ &= \frac{4 \times (22/7)^2 \times 6.4 \times 10^6}{(24 \times 60 \times 60)^2} \\ &= \frac{4 \times 484 \times 64 \times 10^6}{49 \times (24 \times 3600)^2} \\ &= 0.034 \text{ m/s}^2 \\ & \\ & \text{At equator, latitude } \theta = 0^\circ \\ & \therefore \frac{a_c}{g} = \frac{0.034}{9.8} = \frac{1}{288} \\ \end{array}$
(b) Orbital radius of the earth around the sun $(R) = 1.5 \times 10^{11}\ \text{m}$
Time period = 1 yr = 365 day
$\qquad \qquad \qquad \quad = 365 \times 24 \times 60 \times 60\ \text{s} = 3.15 \times 10^{7}\ \text{s}$
Centripetal acceleration $(a_c) = R\omega^2 = \frac{4\pi^2 R}{T^2}$
$\qquad \ \qquad \qquad \qquad \ = \frac{4 \times (22/7)^2 \times 1.5 \times 10^{11}}{(3.15 \times 10^{7})^2} $
$\qquad \qquad \qquad \quad \ \ = 5.97 \times 10^{-3}\ \text{m/s}^2$
$\therefore \quad \frac{a_c}{g} = \frac{5.97 \times 10^{-3}}{9.8} = \frac{1}{1642}$
Given below in Column I are the relations between vectors a, b and c and in Column II are the orientations of a, b and c in the XY-plane. Match the relation in Column I to correct orientations in Column II.
| Column I | Column II |
|---|---|
| (a) $ a + b = c $ | (i)  |
| (b) $ a - c = b $ | (ii) |
| (c) $ b - a = c $ | (iii)  |
| (d) $ a + b + c = 0 $ | (iv)  |
Consider the adjacent diagram in which vectors A and B are corrected by head and tail. Resultant vector C = A + B
(a) from (iv) it is clear that c = a + b
(b) from (iii) c + b = a → a - c = b
(c) from (i) b = a + c → b - a = c
(d) from (ii) -c = a + b → a + b + c = 0
If $|A|=2$ and $|B|=4$, then match the relation in Column I with the angle $\theta$ between $A$ and $B$ in Column II.
| Column I | Column II |
|---|---|
| (a) $\mathbf{A \cdot B} = 0$ | (i) $\theta = 0^\circ$ |
| (b) $\mathbf{A \cdot B} = +8$ | (ii) $\theta = 90^\circ$ |
| (c) $\mathbf{A \cdot B} = 4$ | (iii) $\theta = 180^\circ$ |
| (d) $\mathbf{A \cdot B} = -8$ | (iv) $\theta = 60^\circ$ |
Given $|A| = 2$ and $|B| = 4$:
(a) $A \cdot B = AB \cos \theta = 0$
$\Rightarrow 2 \times 4 \cos \theta = 0$
$\Rightarrow \cos \theta = 0 = \cos 90^\circ$
$\Rightarrow \theta = 90^\circ$
Thus, option (a) matches with option (ii).
(b) $A \cdot B = AB \cos \theta = 8$
$\Rightarrow 2 \times 4 \cos \theta = 8$
$\Rightarrow \cos \theta = 1 = \cos 0^\circ$
$\Rightarrow \theta = 0^\circ$
Thus, option (b) matches with option (i).
(c) $A \cdot B = AB \cos \theta = 4$
$\Rightarrow 2 \times 4 \cos \theta = 4$
$\Rightarrow \cos \theta = \frac{1}{2} = \cos 60^\circ$
$\Rightarrow \theta = 60^\circ$
Thus, option (c) matches with option (iv).
(d) $A \cdot B = AB \cos \theta = -8$
$\Rightarrow 2 \times 4 \cos \theta = -8$
$\Rightarrow \cos \theta = -1 = \cos 180^\circ$
$\Rightarrow \theta = 180^\circ$
Thus, option (d) matches with option (iii).
If $|A|=2$ and $|B|=4$, then match the relations in Column I with the angle $\theta$ between $A$ and $B$ in Column II
| Column I | Column II |
|---|---|
| (a) $\vert \mathbf{A \times B} \vert = 0$ | (i) $\theta = 30^\circ$ |
| (b) $\vert \mathbf{A \times B} \vert = 8$ | (ii) $\theta = 45^\circ$ |
| (c) $\vert \mathbf{A \times B} \vert = 4$ | (iii) $\theta = 90^\circ$ |
| (d) $\vert \mathbf{A \times B} \vert = 4\sqrt{2}$ | (iv) $\theta = 0^\circ$ |
Given $|A| = 2$ and $|B| = 4$
(a) $|A \times B| = AB \sin \theta = 0$
$ \Rightarrow 2 \times 4 \sin \theta = 0 $
$ \Rightarrow \sin \theta = 0 = \sin 0^\circ $
$ \Rightarrow \theta = 0^\circ $
$ \therefore \text{Option (a) matches with option (iv).} $
(b) $|A \times B| = AB \sin \theta = 8$
$ \Rightarrow 2 \times 4 \sin \theta = 8 $
$ \Rightarrow \sin \theta = 1 = \sin 90^\circ $
$ \Rightarrow \theta = 90^\circ $
$ \therefore \text{Option (b) matches with option (iii).} $
(c) $|A \times B| = AB \sin \theta = 4$
$ \Rightarrow 2 \times 4 \sin \theta = 4 $
$ \Rightarrow \sin \theta = \frac{1}{2} = \sin 30^\circ $
$ \Rightarrow \theta = 30^\circ $
$ \therefore \text{Option (c) matches with option (i).} $
(d) $|A \times B| = AB \sin \theta = 4 \sqrt{2}$
$ \Rightarrow 2 \times 4 \sin \theta = 4 \sqrt{2} $
$ \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} = \sin 45^\circ $
$ \Rightarrow \theta = 45^\circ $
$ \therefore \text{Option (d) matches with option (ii).} $