Range is same for the angle if projection \(\theta \text { and } 90-\theta\)

\(\text {Time of flight }=\frac{2 u}{g}\)

Initially velocity has maximum value and at maximum height velocity becomes zero. After that the velocity becomes negative

Unit vector specifies only direction
\(\hat{A}=\frac{\vec{A}}{|\vec{A}|} \)
\(\therefore \hat{i}+\hat{j}=\frac{\hat{l}+\hat{j}}{|\hat{l}+\hat{j}|} \)
\(\hat{i} \text { and } \hat{j} \text { are orthogonal components of vectors. }\)
\(\therefore|\hat{i}+\hat{j}|=\sqrt{1^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}\)

Answers (a), (b) and (c) are all anticlockwiseand answer (d) alone is in the clockwise direction

By following the law of triangular addition, the resultant vector is given by \(\vec R\) = \(\vec A\) + \(\vec B\) as illustrated below.
The magnitude of the resultant vector \(\vec R\) is given by
\(R=|\vec R|=\sqrt{5^2+7^2+2\times 5\times 7\cos 60^o}\)
\(R=\sqrt{25+49+\frac{70\times 1}{2}}=\sqrt{109}\) units

The angle \(\alpha\) between \(\vec R\) and \(\vec A\) is given by
\(\tan\alpha=\frac{B\sin\theta}{A+B\cos\theta}\)
\(\tan\alpha=\frac{7\times\sin60^o}{5+7\cos60^o}=\frac{7\sqrt{3}}{10+7}=\frac{7\sqrt{3}}{17}\) = 0.713
\(\therefore\alpha=35^o\)

(i) Consider two objects A and B moving with uniform velocities V_A and V_B along the same straight tracks but opposite in direction.
\(\overrightarrow { \xrightarrow { { v }_{ A } } } \overleftarrow { \xrightarrow { { v }_{ B } } } \)
(ii) The relative velocity of object A with respect to object B is
\(\overrightarrow { { V }_{ AB } } =\overrightarrow { { V }_{ A } } -(-\overrightarrow { { V }_{ B } } )=\overrightarrow { { V }_{ A } } +\overrightarrow { { V }_{ B } } \)
(iii) The relative velocity of object B with respect to object A is
\(\overrightarrow { { V }_{ BA } } =-\overrightarrow { { V }_{ B } } -\overrightarrow { { V }_{ A } } =-(\overrightarrow { { V }_{ A } } +\overrightarrow { { V }_{ B } } )\)
(iv) Thus, if two objects are moving in opposite directions, the magnitude of relative velocity of one object with respect to other is equal to the sum of magnitude of their velocities.

(i) When two objects A and B are moving with different velocities, then the velocity of one object A with respect to another object B is called relative velocity of object A with respect to B.
(ii) Consider two objects A and B moving with uniform velocities V_A and V_B, as shown, along straight tracks in the same direction \(\overrightarrow { { V }_{ A } } \),\(\overrightarrow { { V }_{ B } } \) , with respect to ground.
(iii) The relative velocity of object A with respect to object B is \(\overrightarrow { { V }_{ AB } } =\overrightarrow { { V }_{ A } } -\overrightarrow { { V }_{ B } } \)
(iv) The relative velocity of object B with respect to object A is \(\overrightarrow { { V }_{ AB } } =\overrightarrow { { V }_{ B } } -\overrightarrow { { V }_{ A } } \)
(v) Thus, if two objects are moving in the same direction, the magnitude of relative velocity of one object with respect to another is equal to the difference in magnitude of two velocities.

(i) Graphically the slope of the position-time graph will give the velocity of the particle.
(ii) At the same time, if velocity-time graph is given, the distance and displacement are determined, by calculating the area under the curve.
Velocity is given by \(\frac { dx }{ dt } =v\)
(iii) Therefore, dx = vdt
By integrating both sides, \(\int _{ { x }_{ 1 } }^{ { x }_{ 2 } }{ dx=\int _{ { x }_{ 1 } }^{ { x }_{ 2 } }{ v\quad dt } } \)
Integration is equivalent to area under the given curve.
(iv) So the term \(\int _{ { t }_{ 1 } }^{ { t }_{ 2 } }{ vdt } \) represents the area under the curve v as a function of time.
(v) Since the left hand side of the integration represents the displacement travelled by the particle from time t₁ to t₂, the area under the velocity time graph will give the displacement of the particle.

Displacement in the velocity - time graph
(vi) If the area is negative, it means that displacement is negative, so the particle has travelled in the negative direction.  

(i) The instantaneous velocity at an instant or simply 'velocity' at an instant t is defined as limiting value of the average velocity as \(\Delta \)t \(\rightarrow\)0, evaluated at time t.
(ii) In other words, velocity is equal to rate of change of position vector with respect to time. Velocity is a vector quantity.
\(\overrightarrow { v } =\lim _{ \Delta x\rightarrow 0 }{ \frac { \Delta \overrightarrow { r } }{ \Delta \overrightarrow { t } } =\frac { d\overrightarrow { r } }{ dt } } \)
(iii) In component form, this velocity is
\(\overrightarrow { v } =\frac { d\overrightarrow { r } }{ dt } =\frac { d }{ dt } (x\hat { i } +y\hat { j } +z\hat { k } )\)
\(=\frac { dx }{ dt } \hat { i } =\frac { dy }{ dt } \hat { j } +\frac { dz }{ dt } \hat { k } \)
Here \(\frac { dx }{ dt } ={ v }_{ x }=x-\) component of velocity.
\(\frac { dx }{ dt } ={ v }_{ y }=y-\) component of velocity.
\(\frac { dz }{ dt } ={ v }_{ z }=z\) - component of velocity.
(iv) The magnitude of velocity v is called speed and is given by
\(v=\sqrt { { v }_{ x }^{ 2 }+{ v }_{ y }^{ 2 }+{ v }_{ z }^{ 2 } } \)
Speed is always a positive scalar. The unit of speed is also meter per second.

(i) The average velocity is defined as ratio of the displacement vector to the corresponding time interval
\(\overrightarrow { { v }_{ avg } } =\frac { \Delta \overrightarrow { r } }{ \Delta t } \)
(ii) It is a vector quantity. The direction of average velocity is in the direction of the displacement vector (\(\Delta \)\(\overrightarrow { r } \)).

(i) If a function is represented by y = f (x), then dy/dx represents the derivative of y with respect to x.
(ii) Mathematically this represents the variation of y with respect to change in x, for various continuous values of x.
(iii) Mathematically the derivative dy/dx is defined as follows
\(\frac { dy }{ dx } =\underset { \Delta x\rightarrow 0 }{ lim } \frac { y(x+\Delta x)-y(x) }{ \Delta x } \)
\(=\lim _{ \Delta x\rightarrow 0 }{ \frac { \Delta y }{ \Delta x } } \)
\(\frac { dy }{ dx } \) represents the limit that the quantity \(\frac { \Delta y }{ \Delta x } \)
attains, as \(\Delta \)x tends to zero. 

(i) In terms of position vector, the displacement vector is given as follows. Consider a particle moving from a point P₁ having position vector \(\overrightarrow { { r }_{ 1 } } ={ x }_{ 1 }\hat { i } +{ y }_{ 1 }\hat { j } +{ z }_{ 1 }\hat { k } \) to a point P₂ where its position vector is \(\overrightarrow { { r }_{ 2 } } ={ x }_{ 2 }\hat { i } +{ y }_{ 2 }\hat { j } +{ z }_{ 2 }\hat { k } \)
(ii) The displacement vector is given by \(\Delta \overrightarrow { r } =\overrightarrow { { r }_{ 2 } } -\overrightarrow { { r }_{ 1 } } \)
= (x₂-x₁)\(\hat { i } \) + (y₂-y₁)\(\hat { j } \)+(z₂-z₁)\(\hat {k } \)
(iii) This displacement is also shown in

(i) If two vectors\(\overrightarrow { A } \) and \(\overrightarrow { B } \) form adjacent sides in a parallelogram, then the magnitude of |\(\overrightarrow { A } \) \(\times\)\(\overrightarrow { B } \)| will give the area of the parallelogram as represented graphically.

 Area of Parallelogram
It divides a parallelogram into two equal triangles as shown. The area of a triangle with \(\overrightarrow { A } \) and \(\overrightarrow { B} \) as sides is \(\frac { 1 }{ 2 } |\overrightarrow { A } \times \overrightarrow { B } |\)

Area of triangle 

360^o = 27 days
1 day = \(\frac{360^o}{27}=13^o.3'\)

ω = \(\frac{\pi}{12}rads^{-1}\) 
θ = ωt
\(=\frac{\pi}{12}\times 4=\frac{\pi}{3}\)
\(= \frac{18}{3}=60^o\)

Angular displacement: While a particle is revolving around a point in a circular path, the angle described by the particle about the axis of rotation (or at the centre of the circle) in a given time is called angular displacement. Its unit is radian.
Angular velocity: The rate of change of angular displacement is called angular velocity. Its unit is rad s^-1
\(\omega =\frac { d\theta }{ dt } \)

Velocity:
Velocity is equal to the rate of change of position vector with respect to time. 
It is a vector quantity \(\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}\)
Speed:
The magnitude of velocity is called speed and is given by \(v= \sqrt{v^2_x+v^2_y+v^2_z}\). It is a positive scalar.

(i) Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object and its direction is from the initial to final position of the object, during the given interval of time. It is a vector quantity.
(ii) Distance is the actual path length travelled by an object in the given interval of time during the motion. It is a positive scalar quantity.

The condition for the two vectors \(\vec{a}\ and \ \vec{b}\) to be perpendicular to each other is \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=0\).
Diagrammatically, when \(\vec{a}\ and \ \vec{b}\) are perpendicular to each other
\(|\vec{a}+\vec{b}| =|\vec{a}-\vec{b}|  \)
\(\text {Squaring }|\vec{a}+\vec{b}|^{2} =|\vec{a}-\vec{b}|^{2}  \)
\(a^{2}+b^{2}+2 a b \cos \theta =a^{2}+b^{2}-2 a b \cos \theta .  \)
\(4 a b \cos \theta =0  \)
\(\text { or } \cos \theta =0  \)
\(\theta =\frac{\pi}{2}\)
If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\) then \(\vec{a}\ and \ \vec{b}\) are perpendicular to each other.

Let v_s and v_r represent the velocities of the swimmer and river respectively with respect to ground
v_s + v_r = 12  ............(1)
and   v_s - v_r = 6  ............ (2) 
Adding the both equations (1) and (2) 2v_s = 12 + 6 = 18 km h^-1 or v_s = 9 km h^-1
From equation (1),
9+ v_r = 12 or v_r = 3 km h^-1
When the river flow and swimmer move in the same direction, the net velocity of swimmer is 12 km h^-1.

(i) Vertical component of initial velocity.
(ii) Time of flight.

\(|3\hat { i } -2\hat { j } +\sqrt { 3 } \hat { k } |=\sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ \sqrt { 3 } }^{ 2 } } \)
\(\sqrt { 9+4+3 } =\sqrt { 16 } =4\)

One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle.
1 rad = \(\frac{180}{\pi}\) degree = 57.295^o