In figure, 0 is the observation point at the ground, A and B are the positions of aircraft for which \(\angle AOB=30°\). Draw a perpendicular OC on AB.Here OC = 3400 m and \(\angle AOC=\angle COB=15°\).Time taken by aircraft from A to B is 10 s.
In \(\Delta \)AOC, AC=OC \(\tan { 15° } \)
= 3400\(\times \)0.2679  
= 910.86 m
AB = AC + CB = AC + AC = 2AC
=2\(\times \)910.86 m
Speed of the aircraft 
v = \(\frac { distance \ AB }{ time } =\frac { 2\times 910.86 }{ 10 } \)
= 182.17 ms^-1 = 182.2 ms^-1

If the dot product of two vectors is zero, then the vectors are orthogonal.

No, it is not a projectile, because a projectile should have two component velocities in two mutually perpendicular directions but in this case, the body has velocity only in one direction while going up or coming down.

Velocity of boat with respect to ground, \({ v }_{ g }=3\overset { \wedge }{ i } +4\overset { \wedge }{ j }\)
Velocity of water with respect to ground,
\({ v }_{ w }=-3\overset { \wedge }{ i } -4\overset { \wedge }{ j } \)
∴ Relative velocity of bat w.r.t water,
\({ v }_{ gw }={ v }_{ g }-{ v }_{ w }=3\overset { \wedge }{ i } +4\overset { \wedge }{ j } -(-3\overset { \wedge }{ i } -4\overset { \wedge }{ j } )= (6\overset { \wedge }{ i } + 8\overset { \wedge }{ j } )\)

Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, \(\theta ={ 45 }^{ o }\)  Then, \({ R }_{ max }={ u }^{ 2 }/g\)
Here,   \({ u }^{ 2 }/g=100m\)
In order to study the motion of the ball along vertical direction, consider a point on the surface of Earth as the origin and vetical upward direction as the positive direction of Y-axis. Taking motion of the ball along vertical upward direction, we have
\({ u }_{ y }=u,{ a }_{ y }=-g,{ v }_{ y }=0,t=?,{ y }_{ o }=0,y=?\)
As,  \({ v }_{ y }={ u }_{ y }+{ a }_{ y }t\)
\(\\ 0=u+(-g)t\Rightarrow t=u/g\)
Also,  \(y={ y }_{ o }+{ u }_{ y }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\)
\(\\ y=0+u(u/g)+\frac { 1 }{ 2 } (-g){ u }^{ 2 }/{ g }^{ 2 }\)
\(\\ =\frac { { u }^{ 2 } }{ g } -\frac { 1 }{ 2 } \frac { { u }^{ 2 } }{ g } =\frac { 1 }{ 2 } \frac { { u }^{ 2 } }{ g } =\frac { 100 }{ 2 } =50m\quad [\frac { { u }^{ 2 } }{ g } =100]\)

Horizontal range,
  R = \(\frac { { u }^{ 2 }sin2\theta }{ g } or\quad 3=\frac { { u }^{ 2 }sin6{ 0 }^{ o } }{ g } =\frac { { u }^{ 2 } }{ g } \sqrt { 3/2 } \)
or   \(\frac { { u }^{ 2 } }{ g } 2\sqrt { 3 } \)
Since, the muzzle velocity is fixed
Therefore, maximum horizontal range,
  \({ R }_{ max }=\frac { { u }^{ 2 } }{ g } 2\sqrt { 3 } =3.464km\)
So, the bullet cannot hit the target.

v = 8km/h, θ = 53⁰7'33''

Here, actual path length travelled, s = 23 km; Displacement = 10 km;
Time taken, t  = 28 min = \(\frac{28}{60} h\)
(a) Average speed of taxi = \(\frac{\text { actual path length }}{\text { time taken }}=\frac{23}{\frac{28}{60}} k \frac{m}{h}=49.3 \mathrm{~km} / \mathrm{h}\)
(b) Magnitude of average velocity = \(=\frac{\text { displacement }}{\text { time taken }}=\frac{10}{\frac{28}{60}} \mathrm{~km} / \mathrm{h}=21.4 \mathrm{~km} / \mathrm{h}\)
The average speed is not equal to the magnitude of average velocity. The two are equal for the motion of taxi along a straight path in one direction.

When a stone is going around a circular path, the instantaneous velocity of stone is acting as tangent to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continues to move along the tangent to circular path. So, the stone flies off tangentially to the circular path.

Area of parallelogram \(|A\times B|=AB\sin\theta =\frac { 1 }{ 2 } AB\)
\(\sin\theta =\frac { 1 }{ 2 } =\sin30^{ 0 }or\ \theta =30^{ 0 } \)

As we know that
\(\left| A+B \right| =\sqrt { A^{ 2 }+{ B }^{ 2 }+2AB\quad cos\quad \theta } \)
And    \(\left| A-B \right| =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\quad \theta } \)  
But as per question, we have
\(\sqrt { A^{ 2 }+{ B }^{ 2 }+AB\quad cos\quad \theta } =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\theta } \)
Squaring both sides, we have (4 AB cos ) = 0
\(\sqrt { A^{ 2 }+{ B }^{ 2 }+AB\quad cos\quad \theta } =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\theta } \)
Hence, the two vectors A and B are perpendicular to each other.

If three vectors acting on a point object at the same time are represented in magnitude and direction by tlte three sides of a triangle taken in the same order, their" resultant is zero.
The object is said to be in equilibrium

A = F, B = F, R = F, θ = ?
\(R=\sqrt { A^{ 2 }+B2+2AB \cos\theta } \)
\(\Rightarrow R^{ 2 }=A^{ 2 }+B2+2AB \cos\theta \)
\(F^{ 2 }=F^{ 2 }+F^{ 2 }+2F^{ 2 }\cos\theta\)
\(1=2(1+\cos\theta ) \)
\( \cos\theta =\frac { 1 }{ 2 } -1=\frac { -1 }{ 2 } =120^{ 0 } \)

For Unit Vector 
\(|\hat { i } +m\hat { j } ++\hat { k } |=\sqrt { 1+m^{ 2 }+1 } =1\)
\(m^{ 2 }+2=1\)
\(m^{ 2 }=-1\Rightarrow m=\sqrt { -1 }\)

Yes, because they cannot be represented by the three sides of a triangle taken in the same order

When the boat is anchored in the harbour, the flag flutters along the N-E direction. It shows that the velocity of wind is along the North-East direction. When the boat starts moving, the flag will flutter along the direction of relative velocity of wind W.r.t. boat. Let v_wb be the relative velocity of wind W.r.t.boat and \(\beta \) be the angle between v_wb and v_w.
Then,    v_wb= v_w + (-v_b) 

Here,  |v_w| = 72 km/h and |-v_b| = 15 km/h
Angle between v_w and - v_b is \(135°\)  i.e. \(\theta =135°\) Then, 
\(\tan { \beta } =\frac { 51\sin { 135° } }{ 72+51\cos { 135° } } =\frac { 51\sin { 45° } }{ 72+51(-\cos { 45°) } } \)
=\(\frac { 51\times (1/\sqrt { 2 } ) }{ 72-51(1/\sqrt { 2 } ) } \) = 1.0039
\(\therefore \)  \(\beta =\tan ^{ -1 }{ (1.0039)=45.1° } \)
Angle w.r.t. East direction = \(45.1°\)-\(45°\)=\(0.1°\)
It means the flag will flutter almost due East.    

Figure shows vectorially the situations,
 
Velocity of rain falling vertically downward v_r = 30 m/s
Velocity of woman riding a bicycle v_w=10 m/s (North to South)
To protect herself from rain, the woman should hold her umbrella in the direction of relative velocityof the rain with respect to the woman i.e.v_rw.
The relative velocity of rain with respect to the woman i.e.
v_rw=v_r-v_w
\(\left| { v }_{ rw } \right| =\sqrt { { (30) }^{ 2 }+{ (10) }^{ 2 } } \)
\(=\sqrt { 900+100 } =\sqrt { 1000 } \)
= \(10\sqrt { 10 } \) m/s
If v_rw makes an angle \(\alpha \) with the vertical, then 
\(\tan { \alpha } =\frac { { v }_{ w } }{ { v }_{ r } } =\frac { OB }{ OA } =\frac { 10 }{ 30 } \)
=\(\frac { 1 }{ 3 } \) = 0.3333
 \(\Rightarrow \alpha =18°26\prime \)
Hence, woman should hold her umbrella at an angle \(18°26\prime \) with the vertical towards South.

Given, R = 1000m
Horizontal range of the bullet fired at an angle \(\theta \) is
\(R=\frac { { u }^{ 2 }sin2\theta }{ g } \Rightarrow 1000=\frac { { u }^{ 2 }sin\theta cos\theta }{ g } ....(i)\)
Bullet is fired from the car moving with 36km/h
i.e.10m/s, then horizontal component of the velocity of = usin\(\theta \) + 10
Vertical component of the velocity of the bullet=usin​\(\theta \) 
Then, new range of the bullet is
\({ R }_{ 1 }=\frac { 2 }{ g } (usin\theta )(ucos\theta +10)\)
\(\\ =\frac { 2 }{ g } { u }^{ 2 }sin\theta cos\theta +\frac { 20 }{ g } usin\theta \Rightarrow { R }_{ 1 }=R+\frac { 20 }{ g } usin\theta\)
\( \\ \Rightarrow { R }_{ 1 }-R=\frac { 20 }{ g } usin\theta ....(ii)\)
\(\\ From\ Eq.(i),we \ have \ u=\sqrt { \frac { 1000\times g }{ 2sin\theta cos\theta } } .....(iii)|\)
\(\\ Now, \ substituting \ the \ value \ of \ u \ in \ Eq.(ii), \ we \ get\ \)
\(\\ { R }_{ 1 }-R=\frac { 20 }{ g } \sqrt { \frac { 1000\times g }{ 2sin\theta cos\theta } } sin\theta =20\sqrt { \frac { 500\times sin\theta }{ gcos\theta } }\)
\( \\ =20\sqrt { \frac { 500 }{ 9.8 } tan\theta } =142.9\sqrt { tan\theta } \) ​​​​

Given, initial velocity (u) = 40m/s
Height of the hall (H) = 25m
Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25m.
Maximum height attained by the ball
\(H=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \Rightarrow 25=\frac { { (40) }^{ 2 }{ sin }^{ 2 }\theta }{ 2\times 9.8 }\)
\(or\ { \sin }^{ 2 }\theta =\frac { 25\times 2\times 9.8 }{ 1600 } =0.3068\)
\(or\ \sin\theta =0.5534=\sin{ 33.6 }^{ 0 }\)
\(or\ \theta ={ 33.6 }^{ 0 }\)
\( \therefore \text{ Horizontal range (R)}=\frac { { u }^{ 2 }\sin2\theta }{ g }\)
\( \\ =\frac { { (40) }^{ 2 }sin2\times { 33.6 }^{ 0 } }{ 9.8 } =\frac { 1600\times sin{ 67.2 }^{ 0 } }{ 9.8 } \)
\(\\ =\frac { 1600\times 0.9219 }{ 9.8 } =150.5m\)

Unit vector perpendicular to both
\(\overrightarrow{\mathrm{A}}=2 \hat{i}+\hat{j}+\hat{k} \text { and } \hat{i}-\hat{j}+2 \hat{k}\)
is given by \(\hat{n}=\frac{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}}{|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|}\)
\(=\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right|\)
\(=\hat{i}[2-(-1)]-\hat{j}(4-1)+\hat{k}(-2-1)\)
\(=3 \hat{i}-3 \hat{j}-3 \hat{k}\)
Unit vector is \(\hat{n}=\frac{3 \hat{i}-3 \hat{j}-3 \hat{k}}{\sqrt{9+9+9}}\)
  \(\frac { 3\hat { i } -\hat { 3j } +3\hat { k } }{ \sqrt { 27 } } \) 

Here, v = 27 kmh^-1 = 27\(\times \)(1000 m) \(\times \)(60\(\times \)60 s)^-1 = 7.5 ms^-1, r = 80 m
Centripetal acceleration, a_c=\(\frac { { v }^{ 2 } }{ r } =\frac { { (7.5) }^{ 2 } }{ 80 } \) = 0.7 ms^-2
Let the cyclist applies the brakes at the point P of the circular turn, then tangential acceleration a_T will act opposite to velocity.
Acceleration along the tangent, a_T = 0.5 ms^-2
Angle between both the accelerations is \(90°\)
Therefore, the magnitude of resultant acceleration \(a=\sqrt { { a }_{ C }^{ 2 }+{ a }_{ T }^{ 2 } } =\sqrt { { (0.7) }^{ 2 }+{ (0.5) }^{ 2 } } \)
   
Let the resultant acceleration make an angle \(\beta \) with the tangent i. e. the direction of net acceleration of the cyclist then, \(\tan { \beta } =\frac { { a }_{ C } }{ { a }_{ T } } =\frac { 0.7 }{ 0.5 } \) = 1.4 or \(\beta =54°28\prime \)  

\(\alpha=\frac{1}{2} \cos ^{-1}\left(1-\frac{u^2}{u^2+g h}\right)=43^{\circ} 47^{\prime}\)

Let the ball strike the nth step of stairs.
∴ Vertical distance travelled \(=\mathrm{ny}=\mathrm{n} \times 0.2=\frac{1}{2} \mathrm{gt}^2\)
Horizontal distance travelled \(=\mathrm{nx}=\mathrm{ut}\)
\(\Rightarrow \mathrm{t}  =\frac{\mathrm{nx}}{\mathrm{u}} \)
\(\mathrm{ny}  =\frac{1}{2} \mathrm{gt}^2=\frac{1}{2} \mathrm{~g}\left(\frac{\mathrm{nx}}{\mathrm{u}}\right)^2=\frac{1}{2} \mathrm{~g} \frac{\mathrm{n}^2 \mathrm{x}^2}{\mathrm{u}^2} \)
\(\Rightarrow \mathrm{n}  =\frac{2 \mathrm{u}^2 \mathrm{y}}{\mathrm{gx^{2 }}} \)
\(=\frac{2 \times 1.8^2 \times 0.2}{9.8 \times 0.2^2} \approx 3.3\)
So, Fourth step.