For a free falling body, initial vertical downward velocity = 0
For a body thrown horizontally with a velocity, initial vertical downward velocity = 0
Since both having same initial vertical downward velocity they cover equal vertical distance at any instant

Velocity and acceleration are in the same direction: So speed increases.

\(v=a t+b t^{2}\)
\(\text { Dimensional equation is } \mathrm{LT}^{-1}\)
\(=a T=b T^{2}\)
\(\therefore \text { The dimension of } b=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^{2}}=\mathrm{LT}^{-3}\)

The motion of the cricket ball in air is essentially a projectile motion. As we have already seen, the range (horizontal distance) of the projectile motion is given by
\(R=\frac{u^2sin2\theta}{g}\)
The initial speed u = 30 ms^-1
The projection angle θ = 30°
The horizontal distance travelled by the cricket ball  \(R=\frac{(30)^2\times sin60^0}{10}=\frac{900\times\frac{\sqrt 3}{2}}{10}=77.94\ m\)
This distance is greater than the distance of the boundary line. Hence the ball will cross this line and go for a six.

From the safe turn condition the speed of the vehicle (v) must be less than or equal to \(\sqrt { { \mu }_{ s }rg } \)
v ≤ \(\sqrt { { \mu }_{ s }rg } \)
\(\sqrt { { \mu }_{ s }rg } \)=\(\sqrt { 0.81\times 10\times 10 } \)=9 ms^-1
For car C, \(\sqrt { { \mu }_{ s }rg } \) is less than v.
The speed of car A, B and Care 7 m s^-1, 8 m s^-1 and 10 m s^-1 respectively. The cars A and B will have safe turns. But the car C has speed 10 m s^-1 while it turns which exceeds the safe turning speed. Hence, the car C will skid.

Kinetic friction comes to playas the block is moving on the surface.
The forces acting on the mass are the normal force perpendicular to surface, downward gravitational force and kinetic friction f_k along the surface.
mg sinθ - fk = ma
But a = g/2
mg sin 60⁰ - f_k = mg/2
\(\frac { \sqrt { 3 } }{ 2 } \)mg - f_k = mg/2
f_k = mg\(\left( \frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } \right) \)
f_k =\(\left( \frac { \sqrt { 3 } -1 }{ 2 } \right) mg\)
There is no motion along the y-direction as normal force is exactly balanced by the mg cos θ
mg cosθ = N = mg/2
f_k = μ_kN = μ_k mg/2
μ_k = \(\frac { \left( \frac { \sqrt { 3 } -1 }{ 2 } \right) mg }{ \frac { mg }{ 2 } } \)
u_k = \(\sqrt { 3 } \)-1

We know that speed is the magnitude of the velocity vector. Hence,
Speed of A = \(|\vec{V_A}|=\sqrt{(3)^2+(-5)^2+(2)^2}=\sqrt{9+25+4}=\sqrt{48}ms^{-1}\)
Speed of B = \(\vec{V_B}=\sqrt{(1)^2+(2)^2+(3)^2}=\sqrt{1+4+9}=\sqrt{14}ms^{-1}\)
Spedd of C = \(\vec{V_C}=\sqrt{(5)^2+(3)^2+(4)^2}=\sqrt{25+9+16}=\sqrt{50}ms^{-1}\)
The particle C has the greatest speed.
\(\sqrt{50}>\sqrt{38}>\sqrt{14}\)

Angle \(\theta\) = 1° 55'= 115' = (115\(\times\)60)"\(\times\) (4.85\(\times\) 10^-6) rad = 3.34\(\times\)10^-2 rad
Since 1" = 4.85\(\times\)10^-6 rad
Radius of the Earth = 6.4\(\times\)10⁶m
From the Figure AB is the diameter of the Earth (b) = 2\(\times\)6.4\(\times\)10⁶m
Distance of the Moon from the Earth x = ?
\(x={b\over \theta}={2\times 6.4 \times 10^6\over 3.34\times 10^{-2}}\)
x = 3.83\(\times\)10⁸m

Mass of a bullet (m) = \(\frac { 40 }{ 1000 } kg\) = 0.04kg
Initial velocity (u) = 250 ms^-1
Final velocity (v) = 0.
displacement (s) = 20cm = 20\(​​\times\)10^-2m
Acceleration a bullet \(a=\frac { { v }^{ 2 }-{ u }^{ 2 } }{ 2s } \)
\(=\frac { { 0 }^{ 2 }-\left( 250 \right) ^{ 2 } }{ 2\times (20\times 10^{ -2 }m) } \)
= \(\frac { -250\times 250 }{ 2\times 20\times { 10 }^{ -2 }m } \)
= \(\frac { -62,50,000 }{ 40\times { 10 }^{ -2 } } \)
Acceleration 'a'  = - 156.25 x10⁵ ms^-2
Now, Average force F = ma
F = 0.04 \(​​\times\)(-156.25 x 10⁵)
= 6.25\(​​\times\)10⁵N

Total mass = 30 kg  m₁ = m₂ = 20 kg
Mass of a third piece, m₃ =Total mass - m₁ and m₂ = 30 kg - 20 kg
Kinetic energy of m₁ and m₂ = 400 J
Kinetic energy of m₃ =?
\({ V }_{ 3 }=\sqrt { \frac { 2\left( { E }_{ 1 }+{ E }_{ 2 } \right) }{ \left( { m }_{ 1 }+{ m }_{ 2 } \right) } } \)
\({ V }_{ 3 }=\sqrt { \frac { 2\times 400 }{ 20 } } \)
\({ V }_{ 3 }=\sqrt { \frac { 800 }{ 20 } } \)\(=\sqrt { 40 } \)
V₃ = 6.32 ms^-2
Let V₃ be the speed of the Kg mass

(i) Consider a well without water, of some depth d. Take a small object (for example lemon) and a stopwatch. When you drop the lemon, start the stopwatch. As soon as the lemon touches the bottom of the well, stop the watch. Note the time taken by the lemon to reach the bottom and denote the time as t.
(ii) Since the initial velocity of lemon u=0 and the acceleration due to gravity g is constant over the well, by using use the equations of motion for constant acceleration.
\(s=ut+\frac { 1 }{ 2 } { a }t^{ 2 }\)
(iii) u = 0, s = d, a = g (Since we choose the y axis downwards), Then
\(d=\frac { 1 }{ 2 } { gt }^{ 2 }\)
(iv) Substituting g = 9.8 ms^-2 the depth of the well is obtained.
(v) To estimate the error in calculation another method is used to measure the depth of the well.
(vi) Take a long rope and hang the rope inside the well till it touches the bottom. Measure the length of the rope which is the correct depth of the well (d_correct). Then
error = d_correct - d
relative error = \(\frac { { d }_{ correct }-d }{ { d }_{ correct } } \)
percentage of relative error
\(=\frac { { d }_{ correct }-d }{ { d }_{ correct } } \times 100\) 

Mass of the ball (m) = 100 g = 0.1 kg
Speed of the ball (v) = 15 ms^-1
Velocity of the ball after rebounding,
\({ V }^{ 1 }=-\frac { V }{ 2 } =-\frac { 15 }{ 2 } m/s\)
\({ V }^{ 1 }=-7.5m/s^{ -1 }\)
Change in momentum =m(v¹ - v)
= 0.1\(\times\)(-7.5 -15)
Change in momentum = -22.5 Ns

(i) Newton's second law gives the correct explanation for the experiment on the inclined plane.
(ii) In normal cases, where friction is not negligible, once the object reaches the bottom of the inclined plane, it travels some distance and stops.
(iii) Note that it stops because there is a frictional force acting in the direction opposite to its velocity.
(iv) It is this frictional force that reduces the velocity of the object to zero and brings it to rest.
(v) As per Aristotle's idea, as soon as the body reaches the bottom of the plane, it can travel only a small distance and stops because there is no force acting on the object.

(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) 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.  

3.08 x 10¹⁶m equivalent to 1 parsec
1 metre is equivalent to \({{1}\over{3.08\times{10}^{16}}}\)
= 0.324\(\times\)10^-16
= 3.24\(\times\)10^-17 m
In one metre 3.24\(\times\)10^-17 parsec are present.

(i) Tangential acceleration = g sin ፀ
(ii) mg cos ፀ acts along OP (outwards)
(iii) tension (T) acts along PO (inwards)
Net force on the body at P acting along
PO = T- mg cos ፀ

This must provide the necessary centripetal force. \(\frac{mv^2}{r}\)
\(\therefore T -mg \ cos \ \theta=\frac{mv^2}{r} \)  
Centripetal acceleration (a_⊥) =  \(\frac { T-mg \ cos\theta }{ m } \)

\(\overrightarrow { r } =3\hat { i } +2\hat { j } \)
\(\text { Unit vector } =\frac{\vec{r}}{|\vec{r}|}  \)
\(|\vec{r}| =\sqrt{3^{2}+2^{2}}=\sqrt{13}  \)
\(\therefore \text { Unit vector } =\frac{3 \hat{i}+2 \hat{j}}{\sqrt{13}}\)

The speed of sound in water v = 1460 ms^-1
Time taken t = 80 s
\(=\frac{2 d}{t} \)
2d = u x t 
= 1460 x 80 
2d =11680 m
\(\therefore \text { Distance of enemy submarine } d =\frac{11680}{2} \)
= 58.40 km

(i) Physical quantities are classified into two types. There are fundamental and derived quantities.
(ii) Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities.
(iii) These are length, mass, time, electric current, temperature, luminous intensity and amount of substance.
(iv) Quantities that can be expressed in terms of fundamental quantities are called derived quantities.
(v) For example, area, volume, velocity, acceleration, force.

The echo will be heard when the sound reaches back at the place of firing. So, the total distance travelled by sound is 2\(\times\)1.2 km = 2.4 km = 2400 m.
\(speed=\frac{2400m}{8s}=300\ ms^{-1}\)

The centripetal acceleration ac=\(\frac { { v }^{ 2 } }{ r } \)
V.- velocity of Earth around the orbit
r - radius of orbit or distance of Earth to Sun
Velocity of Earth is written in terms of angular velocity (ω) as
v-ωr
By substituting in the centripetal acceleration formula, ac=\({ a }_{ c }=\frac { { \omega }^{ 2 }{ r }^{ 2 } }{ r } =\omega ^{ 2 }\)=ω²r
But ω=\(\frac { 2\pi }{ T } \) where T is time for the Earth to orbit around the sun, which is one year.
T = 365 days = 365\(\times\)24\(\times\)60\(\times\)60s
T = 31\(\times\)10⁷ s
(ω)= 2.02\(\times\)10^-7 rad per see
a_c = (2.02\(\times\)10^-7)² \(\times\)(150\(\times\)10⁸)
a_c= 6.12\(\times\)10^-4 ms^-2

In cgs system 76 cm of mercury pressure = 76\(\times\)13.6\(\times\)980 dyne cm^-2
The dimensional formula of pressure P is [ML^-1T^-2]; so P₁ [M₁^a L₁^b T₁^c ]= P₂[M₂^a L₂^bT₂^c]
We have P₂ = P₁ \([{M_1\over M_2}]^a[{L_1\over 2}]^b[{T_1\over T_2} ]^c\)
M₁ = 1g, M₂ =1kg
L₁ = 1 cm, L₂ = 1m
T₁ = 1 s, T₂ = 1s
So, a = 1, b = -1, and c =-2
Then, P₁ = 76\(\times\)13.6\(\times\)980
\([{1g\over 1kg}]^1[{1cm\over 1m}]^{-1}[{1s\over 1s}]^{-1}=76\times13.6\times980[{10^{-3}kg\over 1kg}]^1[{10^{-2}m\over1m}]^{-1}[{1s\over 1s}]^{-2}\)
= 76\(\times\)13.6\(\times\)980\(\times\)[10^-3]\(\times\)10²
P₂ = 1.01\(\times\)10⁵ Nm^-2

(i) Gravitational force (mg) acting downwards
(ii) Normal force (mg) acting upwards
(iii) Frictional force (F_s) acting horizontally Inwards along the road.

Downward gravitational force (mg), Normal force perpendicular to inclined surface (N).

Since the acceleration is the second derivative of position vector of the body
\(\left( \overset { \rightarrow }{ a } =\frac { { d }^{ 2 }\overset { \rightarrow }{ r } }{ { dt }^{ 2 } } \right) \) , the force on the body is
\(\overset { \rightarrow }{ F } =m\frac { { d }^{ 2 }\overset { \rightarrow }{ r } }{ { dt }^{ 2 } } \)

(i) The car is moving with uniform velocity v with respect to a person standing (at rest) on the ground.
(ii) As the car is moving with constant velocity with respect to ground to the person is at rest on the ground, both frames (with respect to the car and to the ground) are inertial frames.

Speed of a sound in water 1450 ms^-1
Time delay T = 110.3s
Distance of the enemy ship, D \(={{\sqrt{T}}\over{2}}\)
\(={{1450\times110.3}\over{}2}\)
\(={{1159935}\over{2}}\)
= 79967 m
Distance of enemy ship = 79.86 km.

Distance, D = 824.7\(\times\)10⁶ km
\(\theta=35.72"={{35.72}\over{60\times 60}}\times{{\pi}\over{180}}\) rad
\(\theta \ \rightarrow\) is angular diameter
Diameter, d = ?
\(d=D\times\theta\)
\(=824.7\times{10}^{6}\times{{35.72}\over{60\times 60}}\times{{\pi}\over{180}}\) km 
\(=824.7\times{10}^{6}\times{{35.72}\over{3600}}\times{{\pi}\over{180}}\) km
\(=824.7\times{10}^{6}\times{{35.72\times3.14}\over{3600\times180}}\)km
Diameter of Jupiter = 1.427\(\times\)10⁵ km

(i) Fractional error \(=\delta_t={{\triangle t}\over{t}}\)
\(={{0.01}\over{5}}=0.002.\)
(ii) Percentage error \(=\delta_t={{\triangle t}\over{ t}}\times100\%\)
= 0.002\(\times\)100%
= 0.02%

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.

(i) Any physical quantity is represented by a "function" in mathematics. For example:- temperature T. We know that the temperature varies with time.
(ii) Mathematically this variation can be represented by the notation 'T (t)' and it should be called "temperature as a function of time".
(iii) It implies that if the value of 't' is given, then the function "T (t)" will give the value of the temperature at that time 't'.

(i) The Distance travelled by an object in motion in a given time is never negative or zero, it is always positive.
(ii) The displacement of an object, in a given time can be positive, zero or negative.
(iii) The displacement of an object can be equal or less than the distance travelled but never greater than distance travelled.
(iv) The distance covered by an object between two positions can have many values, but the displacement between them has only one value (in magnitude).

A coordinate system and the position of an object is described relative to it, then such a coordinate system is called frame of reference.

T = mg
T = 50g x 9.8m/s²
Tension (T) = 0.49 N

According to triangle law of vectors.
\(R =\sqrt{A^{2}+B^{2}+2 A B \cos \theta} \text { and }  \)
\(\tan \alpha =\frac{B \sin \theta}{A+B \cos \theta}  \)
\(\alpha =90^{\circ}(\text { given })  \)
\(\therefore \frac{B \sin \theta}{A+B \cos \theta} =\alpha  \)

\(\therefore A+B \cos \theta =0  \)
\(\text { or } A =-B \cos \theta  \)
\(R =\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
\(=\sqrt{B^{2} \cos ^{2}} \theta+B^{2}-2 B^{2} \cos ^{2} \theta\)
\(=\sqrt{B^{2}-B^{2} \cos ^{2} \theta}  \)
\(=\sqrt{B^{2}\left(1-\cos ^{2} \theta\right)}  \)
\(=\sqrt{B^{2} \sin ^{2} \theta}  \)
\(R =B \sin \theta \)
\(But \ R=\frac{B}{2} (given)\)
\(\therefore \frac{B}{2}=B \sin \theta \)
\(\sin \theta=\frac{1}{2} or \theta=30^{\circ}\)
\(\therefore\) angle between A and B= \(180^{\circ}-\theta=180^{\circ}-30^{\circ}=150^{\circ}\)

Max.height = \(\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \)
\(Range \ R=\frac { { u }^{ 2 }{ sin 2 }\theta }{ 2g } \)
Here, R = 4 x Max.height
\(\therefore \frac { { u }^{ 2 }{ sin 2 }\theta }{ g } =\frac { {4\times u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \)
\(sin2\theta =2{ sin }^{ 2 }\theta \)
\(2sin\theta cos\theta =2{ sin }^{ 2 }\theta \)
\(\therefore \ cos\theta = sin \theta\)
\(\therefore \theta =45°\)

\(\overrightarrow { A } =4\hat { i } -2\hat { j } +\hat { k } \)
\(\overrightarrow { B } =5\hat { i } +3\hat { j } -4\hat { k } \)
Resultant vector = \(\overrightarrow { A } +\overrightarrow { B } \)
\(=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & -2 & 1 \\ 5 & 3 & -4 \end{matrix} \right| \)
\(=\hat{i}(8-3)+\hat{j}[5-(-16)]+\hat{k}(12+10)\)
Resultant vector = 5\(\hat { i } \) +21\(\hat { j } \) + 22\(\hat { k } \)

Least count of screw gauge
\(={{Pitch}\over{No. of\ divisions}}\)
\(={{1}\over{100}}=0.01\ mm\) (or) 0.001 cm
Least count of vernier calipers
\(=1MSD-1-1VSD=(1-19/20)MSD\)
\(={{1}\over{20}}\) = 0.05 cm.
So screw gauge is more precise than vernier.

Radius of the circle r = 3.12 m
Area of the circle A = \(\pi\) r²
= 3.14\(\times\)3.12\(\times\)3.12
= 30.566016 m²
According to the rule of significant 
A = 30.6 m² 

C is the centre of the Earth. A and B are two diametrically opposite places on the surface of the Earth. From A and B, the parallaxes θ₁ and θ₂ respectively of Moon M with respect to some distant star are determined with the help of an astronomical telescope. Thus, the total parallax of the Moon subtended on Earth.
\(\angle AMB=\theta_1+ \theta_2=\theta\)
If θ is measured in radians, then
\(\theta=\frac{A B}{A M} ; A M \approx M C \quad {\theta}=\frac{A B}{M C} \text { or } M C=\frac{A B}{\theta}\)
Knowing the values of AB and θ, we can calculate the distance MC of Moon from the Earth.

F ∝ m^a v^b r^c ;
F = k m^a v^b r^c
where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k= 1, we have
[MLT^-2] = [M]^a [LT^-1]^b [L]^c
= [M^aL^bT^-bL^c]
[MLT^-2] = [M^aL^b+cT^-b]
Comparing the powers of M, L and T on both sides
 a = 1 ; b + c = 1; -b =-2
2 + c = 1 ; b = 2;
a = 1 b = 2 and c = -1
From the above equation we get
F = m^av^br^c
F =m¹v²r^-1
or F = \(\frac { m{ v }^{ 2 } }{ r } \)

Let G_SI be the gravitational constant in the SI system and G_cgs in the cgs system. Then 
G_SI = 6.6 10^-11 Nm² kg^-2;
G_cgs = ?
n₂ =\(n_1{ \left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { L_{ 1 } }{ L_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { T_{ 1 } }{ T_{ 2 } } \right] }_{ }^{ c }\)
G_cgs = G_SI \({ \left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { L_{ 1 } }{ L_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { T_{ 1 } }{ T_{ 2 } } \right] }_{ }^{ c }\)
M₁= 1 kg L₁ = 1 m T₁= 1s
M₂= 1 kg L₂ = 1 m T₂= 1s
The dimensional formula for G is M^-1L³ T^-2
a = -1 b = 3 and c =-2
G_cgs = 6.6\(\times\)10^-11 \(\left[ \frac { 1kg }{ 1g } \right] ^{ -1 }\left[ \frac { 1m }{ 1cm } \right] ^{ 3 }\left[ \frac { 1s }{ 1s } \right] ^{ -2 }\)
=  6.6\(\times\)10^-11 \(\left[ \frac { 1kg }{ { 10 }^{ -3 }kg } \right] ^{ -1 }\left[ \frac { 1m }{ { 10 }^{ -2 }m } \right] ^{ 3 }\left[ \frac { 1s }{ 1s } \right] ^{ -2 }\)
= 6.6\(\times\)10^-11\(\times\)10^-3\(\times\)10⁶\(\times\)1
G_cgs = 6.6\(\times\)10^-8 dyne cm² g^-2

Let us consider two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\)

(i) Represent the vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) by the two adjacent sides of a triangle taken in the same order.
(ii) Then the resultant is given by the third side of the triangle taken in the opposite order.

(iii) The head of the first vector \(\overrightarrow{A}\) is connected to the tail of the second vector \(\overrightarrow{B}.\) 
(iv) Let 9 be the angle between \(\overrightarrow{A}\) and \(\overrightarrow{B}.\) Then R is the resultant vector connecting the tail of the first vector A to the head of the second vector \(\overrightarrow{B}.\)
(v) The magnitude of \(\overrightarrow{R}\) (resultant) is given geometrically by the length of \(\overrightarrow{R}\) (OQ) and the direction of the resultant vector is the angle between \(\overrightarrow{R}\) and \(\overrightarrow{A}\)
(vi) Thus we write \(\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}\)
\(\because \overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ}\)
Magnitude of resultant vector
consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

\(\cos\ \theta={{AN}\over{B}}\therefore AN=B\ \cos\theta\) and
\(\sin \ \theta={{BN}\over{B}}\)  \(\therefore BN=B\sin\theta\)
For L10BN,we have OB2+ ON2+ BN2
\(\Rightarrow\) R² = (A + B cos\(\theta\))² + (B sin\(\theta\))²
\(\Rightarrow\) R² = A²+ B²cos²\(\theta\) + 2AB cos\(\theta\) + B2sin2\(\theta\)
\(\Rightarrow\) R² = A²+ B² (cos2\(\theta\)+ sin2\(\theta\))+ 2AB cos\(\theta\)
\(\Rightarrow\ R=\sqrt{{A}^{2}+{B}^{2}+2AB\cos\ \theta}\)
which is the magnitude of the resultant of \(\overrightarrow{A}\) and \(\overrightarrow{B}.\)
Direction of resultant vectors: If \(\theta\) is the angle between \(\overrightarrow{A}\) and \(\overrightarrow{B}\), then
\(|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{A^2+B^2+2AB\cos\theta}\)
If \(\overrightarrow{R}\) makes an angle \(\alpha\) with \(\overrightarrow{A},\) then in \(\triangle OBN,\)
\(\tan\ \alpha={{CN}\over{ON}}={{CN}\over{OA+AN}}\)
\(\tan\ \alpha={{BN}\over{ON}}={{BN}\over{OA+AN}}\)
\(\tan\ \alpha=\left({{B\sin\theta}\over{A+B\cos\theta}} \right)\)
\(\Rightarrow\alpha = {tan}^{-1}\left( {{B\sin\theta}\over{A+B\cos\theta}} \right)\)

(i) To draw the free body diagram, the block is assumed to be a point mass. Since the motion is on the inclined surface, we have to choose the coordinate system parallel to the inclined surface as shown in Figure (b).
(ii) The gravitational force mg is resolved in to parallel component mg sin e along the inclined plane and perpendicular component mg cos e perpendicular to the inclined surface. (Figure (b)).
(iii) Note that the angle made by the gravitational force (mg) with the perpendicular to the surface is equal to the angle of inclination angle e as shown in Figure (c).

(a)Free body diagram
(b) mg resolved into parallel and perpendicular components

In the Triangle ABC  Total Angle = 90 + \(\theta +{ \theta }_{ 1 }\) = 180 From the above equation
\({ \theta }_{ 1 }\) = 180 -90 - \(\theta\) = 90 - \(\theta\)
But from the figure \({ \theta }_{ 2 }\) =90 - \({ \theta }_{ 1 }\) = 90 -(90-\(\theta\))
It given \({ \theta }_{ 2 }\) =  \(\theta\)
C) The angle \({ \theta }_{ 2 }\) is equal to \(\theta\)
There is no motion(acceleration) along they axis. Applying Newton's second law in the y-direction
-mg cos  \(\theta\)  \(\hat { j } +N\vec { j } \) = 0(No acceleration) 
By comparing the components on both sides, N - mg cos \(\theta =0\)
N=mg cos \(\theta\) 
The magnitude of normal force (N) exerted by the surface is equivalent to mg cos \(\theta\)
(v) The object slides (with an acceleration) along the x-direction. Applying Newton's second law in the x-direction
mg sin \(\quad \theta \hat { i } =ma\hat { i } \)
By comparing the components on both sides, we can equate. mg sin \(\theta\) = ma
The acceleration of the sliding object is a = g sin \(\theta\)
(vi) Note that the acceleration depends on the angle of inclination.\(\theta\)  If the angle 9 is 90 degree, the block will move vertically with acceleration a = g
(vii) Newton's kinematic equation is used to find the speed of the object when it reaches the- bottom. The acceleration is constant throughout the motion
v² = u² + 2as along the x-direction  .......(1)
The acceleration a is equal to mg sin \(\theta\) The initial speed (u) is equal to zero as it starts from rest. Here s is the length of the inclined surface.
The speed (v) when it reaches the bottom is (using equation (1))
\(v=\sqrt { 2sg\ sin\ \theta } \)

(i) The centripetal acceleration is given by \(a=\frac { { v }^{ 2 } }{ r } \) This expression explicitly depends on Moon's speed which is non trivial. We can work with the formula
ω²R_m = a_m 
(ii) a_m is centripetal acceleration of the Moon due to Earth's gravity.
ω is angular velocity
(iii) R_m the distance between Earth and the Moon, which is 60 times the radius of the Earth.
R_m = 60R = 60 \(\times\)6.4 \(\times\)10⁶ = 384 \(\times\)10⁶ m
(iv) As we know the angular velocity ω = \(\frac { 2\pi }{ T } \)and
T = 27.3 days = 27.3 \(\times\) 24 \(\times\) 60 \(\times\) 60 second
=  2.358 \(\times\) 10⁶ sec
(v) By substituting these values in the formula for acceleration 
a_m = \(\frac { (4\pi ^{ 2 })(384\times { 10 }^{ 6 }) }{ (2.358\times { 10 }^{ 6 }) } \)  = 0.00272 ms^-2
= 2.72 x 10^-3ms^-2
The centripetal acceleration of Moon towards the Earth is 0.00272 m s^-2