Class 11 Kinematics Notes

Unit 1- Mechanics

Chapter 3 – kinematics

Static:  The branch of mechanics which deals with body at rest is called static.

Kinematics

The branch of mechanics which deals with the study of the motion of object without taking into the account of the cause of the motion in objects.

Dynamics

 The branch of mechanics which deals with the study of motion of the objects with taking into the account of the cause of the motion in the objects.

Displacement and Distance

The shortest distance between two point is called displacement.

  The length of actual path travelled by body between two point is called distance.

Q. Can a body have zero displacement  but non-zero distance travelled?

Ans: Yes, it can be a body returns to same position after motion, then displacement is zero but distance travelled is non-zero.

Speed and Velocity

The rate of change of displacement of a body is called velocity.It is a vector quantity and its SI unit is m/s.

                                                            Velocity = displacement / time taken

The rate of distance travelled by a body is called speed.It is scalar quantity and it’s SI unit is m/s.

                                                            Speed = distance/time taken

Average speed and instantaneous speed

Average Speed: It is defined as the ratio of total distance travelled  to the  total time taken.

                                                            Average speed = Total distance travelled / total time taken

Instantaneous Speed: The speed of a body at a particular instant of time on it‘s path.

Average velocity and instantaneous Velocity

Average velocity: It is defined as the ratio of total displacement to the total time taken.

                                Average speed = Total displacement / total time taken

Instantaneous velocity: The velocity of a body at a particular instant of time on it‘s path.

Acceleration and Retardation

The rate of change of velocity is called acceleration. It is vector quantity and it’s SI unit is ms^-2

                  a = change in velocity / time

The rate of decrease of velocity is called retardation or negative acceleration.

# Equation of motion with uniform accleration:
1) V = u + at
2) S = $ut+\frac{1}{2}a{t}^2$
3) $V^2 =U^2+2as$
4)$S_{nth}=u+a\left(\frac{2n–1}{2}\right)$
#Equation of motion under gravity:
1) V=$u\pm gt$
2) h=$ut\pm \frac{1}{2}gt$
3)$v^2=u^2\pm 2gh$
4)$S_{nth}=u\pm g\left(\frac{2n–1}{2}\right)$
#Distance travelled in nth second:
we have,
$S_n=un+\frac{1}{2}a{n}^2$
$S_{n–1}=u(n–1)+\frac{1}{2}a(n–1|{)}^2$
Now,
$S{n}^{th}=S_n–S_{n–1}$
$=un+\frac{1}{2}a{n}^2–u\left(n–1\right)–\frac{1}{2}a(n–1{)}^2$
$=un+\frac{1}{2}a{n}^2–un+u–\frac{1}{2}a(n–1{)}^2$
$=u+\frac{1}{2}a[n^2–(n–1{)}^2]$
$=u+\frac{1}{2}a[n^2–n^2+2n–1]$
$=u+a\left(\frac{2n–1}{2}\right)$
$\therefore S_{nth}=u+a\left(\frac{2n–1}{2}\right)$

#Equation of motion (Graphical Treatment)

Let us consider a body is moving with constant acceleration ‘a’ along a straight line AB with intial velocity u(t=0) and u after time ‘t’ it’s final velocity become ‘v’

from figure
OA=ED=U
EB=OC=V
AD=OE=t
[1]. V = U + at
Now, Acceleration of body = Slope of line AB
or, $a=\frac{BD}{AD}=\frac{EB–ED}{AD}$
or,$a=\frac{V–U}{t}$
or, $V–U=at$
$\therefore V=U+at$
[2]. S = $ut + \frac{1}{2}a{t}^2$
we have,
a=$\frac{BD}{AD}=\frac{BD}{t}$
$\therefore BD=at$
Displacement(s) = Area of trapezium of OABE
= Area of Triangle ADB + Area of rectangle OADE
= $\frac{1}{2}(BD\times AD)+AD\times ED$
= $\frac{1}{2}(at\times t)+tu$
$\therefore S=ut+\frac{1}{2}a{t}^2$
[3].$v^2=u^2+2as$
we have,
$\mathrm{a}=\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{EB}–\mathrm{ED}}{\mathrm{AD}}=\frac{\mathrm{EB}–\mathrm{ED}}{\mathrm{AB}}$
$\mathrm{or}, \mathrm{AD}=\frac{\mathrm{EB}–\mathrm{ED}}{\mathrm{AB}}$
now, Displacement(s)= Area of Trapezium of OABE
$\mathrm{or}, \mathrm{s}=\frac{1}{2}(\mathrm{EB}+\mathrm{OA})\times \mathrm{AD}$
$\mathrm{or}, \mathrm{s}=\frac{1}{2}\left(\mathrm{EB}+\mathrm{OA}\right)\times \frac{\mathrm{EB}–\mathrm{ED}}{\mathrm{AB}}$
$or, s=\frac{1}{2}(v+u)\frac{(v–u)}{a}$
$or, s=\frac{1}{2a}(v^2–u^2)$
$or, v^2–u^2=2as$
$\therefore , v^2=u^2+2as$

Numericals

Q.1[A] An object is dropped from the top of the tower height 156.8m and at the same time another object is thrown velocity upward with the velocity of 78.1 m/s from the foot of the tower when and where the object meet?

let A be the top of tower and B it’s foot Also let C be the point where the both object meet .

Now, for the dropped object,
distance travelled
$s=ut+\frac{1}{2}g{t}^2$
$or, x=\frac{1}{2}g{t}^2⇢\left(i\right)$
for the object thrown vertically upward, distance travel,
$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$
$\mathrm{or}, \mathrm{h}–\mathrm{x}=78.1\mathrm{t}–\frac{1}{2}{\mathrm{gt}}^2⇢\left(\mathrm{ii}\right)$
$\mathrm{u}sin\mathrm{g} \mathrm{equation} \mathrm{i} \mathrm{and} \mathrm{ii}$
$\mathrm{or}, \mathrm{h}–\frac{1}{2}{\mathrm{gt}}^2=78.1\mathrm{t}–\frac{1}{2}{\mathrm{gt}}^2$
$\mathrm{or} \mathrm{h}=78.1\mathrm{t}$
$\mathrm{or}, \mathrm{t}=\frac{156.8}{78.1}=2sec$
$\mathrm{putting} \mathrm{t}= 2 sec \mathrm{in} \mathrm{equation} \mathrm{i}$
$\mathrm{we} \mathrm{get},$
$\mathrm{x}=\frac{1}{2}10\times (2{)}^2$
$\mathrm{x} =20\mathrm{m}$
Hemce, they meet 20m below from the top of tower after 2 sec.

Q.1[B] A ball is dropped from the top of a tower 300m height after 1 sec another ball is dropped with 20m/s from the top of the tower. When and where they meet?

Let A be the top of the tower and B be the foot. Also let C be the point where the both object meet.

Now, Case I
$\mathrm{x}=\frac{1}{2}{\mathrm{gt}}^2⇢\mathrm{i}$
then, case ii
$\mathrm{x}=20(\mathrm{t}–1)+\frac{1}{2}\mathrm{g}(\mathrm{t}–1{)}^2⇢\mathrm{ii}$
from the equation i and ii
$or, \frac{1}{2}g{t}^2=20(t–1)+\frac{1}{2}g(t–1{)}^2$
$\mathrm{or}, 5{\mathrm{t}}^2=20+–20+5({\mathrm{t}}^2–2\mathrm{t}+1)$
$or, 5t^2=20t–20t+5t^2–10t+5$
$or, 15=10t$
$\Rightarrow t=1.5 sec$
$putting the value of‘t‘ in eaution i we get;$
$or, x=\frac{1}{2}10\times (1.5{)}^2$
$\therefore , x=11.25$
Hence they meet 11.25m below from the top of after 1.5 sec

Q.1[C] A car travelling with the speed of 15m/s is braked and it slows down with uniform retardation.It covers a distance of 88m and it’s velocity reduces to 7m/s. If the car continuous to slow down with the rate after what further distance will it be broughted to rest?

By the figure,
In case I
u= 15 m/s
s= 88m
v= 7m/s
we have,
$v^2=u^2+2as$
$or, 7^2=(15{)}^2+2as$
$or, 49–225=2\times a\times 88$
$or, a=–1$
$\therefore retardation =1m/s^{\mathit{2}}$
Again case II
U= 7m/s
s =?
v= 0m/s
a= $–1\mathrm{m}/{\mathrm{s}}^2$
we have,
${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$
$\mathrm{or}, 0^2=7^2+2(–1)\times \mathrm{s}$
$\mathrm{or}, 2\mathrm{s} =49$
$\therefore \mathrm{s}=24.5\mathrm{m}$
Hence a car will be broughted to rest after further distance 24.5m.

1Q.[D] A balls falls freely from top of thhe tower and during the last second of it’s fall. It falls through 25m. Find the height of tower?

solution,
In last second it falls through ${\mathrm{St}}^{\mathrm{th}}$=25m
inital velocity(u)= 0m/s
Now, Distance travelled in ${\mathrm{t}}^{\mathrm{th}}\sec \mathrm{ond}$
${\mathrm{St}}^{\mathrm{th}}= \mathrm{u}+\frac{\mathrm{g}}{2}(2\mathrm{t}–1)$
$\mathrm{or}, 25=0+\frac{10}{2}(2\mathrm{t}–1)$
$\mathrm{or},\frac{25}{5}=2\mathrm{t}–1$
$\mathrm{or}, 6=2\mathrm{t}$
$\Rightarrow \mathrm{t} =3sec$
$\mathrm{Again},$
$\mathrm{h}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$
$= 0\times 3+\frac{1}{2}\times 10(3{)}^2$
$= 5\left(9\right)$
$\mathrm{h}= 45\mathrm{m}$
hence the height of the tower is 45m

Q.1[E] if the displacement of body is proportional to the square of time. State the nature of the motion of thhe body?

By question;
$\mathrm{y}\mathrm{\alpha }{\mathrm{t}}^2$

$\mathrm{Now}, \mathrm{velocity}=\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}\left({\mathrm{kt}}^2\right)}{\mathrm{dt}}=2\mathrm{kt}$
then,
$\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}\left(2\mathrm{kt}\right)}{\mathrm{dt}}=2\mathrm{k}\frac{\mathrm{d}\left(\mathrm{t}\right)}{\mathrm{dt}}=2\mathrm{k}$
$\Rightarrow \mathrm{a}=2\mathrm{k}$
Hence, the body moves with constant accleration.

Projectile Motion:

Any body thrown toward space so that it moves only under the action of gravity is called projectile motion. for eg: 1. stone thrown horizonatally 2. A bomb drop from aeroplane

• Ux = U cosθ along horizantal and
• Uy = U sinθ along vertical.

Let us consider a projectile thrown at certain angle with horizantal. Let us consider a projectile thrown towards sky from ground with intial velocity ‘v’ making angle ‘θ’ wih ground. Now it velocity can be resuded into two components. ucosθ along horizantal and usinθ along vertical. since accleration due to gravity act in vertical direction only so, horizantal velocity remains constant but vertical velocity is variable . Since p(x,y) be any point at which projectile takes time’t’ to reach ground.

Motion along horizantal,
$using S=ut+\frac{1}{2}g{t}^2,we get$
$or, x=u_{x}t$
$or, x=ucos\theta .t$
$\therefore t=\frac{x}{ucos\theta }⇢\left(i\right)$
Motion along vertical,
$using S=ut+\frac{1}{2}a{t}^2,we get$
$or, y=u_{y}t+\frac{1}{2}(–g)t^2$
$or, y=usin\theta .t–\frac{1}{2}g{t}^2⇢ii$
using eq i and ii
$or, y=usin\theta \frac{x}{ucos\theta }–\frac{1}{2}.g(\frac{x}{ucos\theta }{)}^2$
$or, y=tanx–\frac{g}{2u^2{\cos }^2\theta }.x^2⇢iii$
$which is the equation of from y=ax+b{x}^2$
$wtih a=tan\theta and b=\frac{–g}{2u^2{\cos }^2\theta } e{q}^n of parabola$
Hence, the path of projectlie is parabolic

Time of flight (T): The time for which projectile remains in space is called time of flight.

since, the projectile returns to ground after time T.
Now,
$h=u_{y}T–\frac{1}{2}g{T}^2$
$\mathrm{or}, 0=\mathrm{u}sin\mathrm{\theta }\mathrm{T}–\frac{1}{2}{\mathrm{gT}}^2$
$\mathrm{or}, \frac{1}{2}{\mathrm{gT}}^2=\mathrm{u}sin\mathrm{\theta }\mathrm{T}$
$\therefore \mathrm{T}=\frac{2\mathrm{u}sin\mathrm{\theta }}{\mathrm{g}} \mathrm{This} \mathrm{is} \mathrm{required} {\mathrm{eq}}^{\mathrm{n}} \mathrm{for} \mathrm{time} \mathrm{of} \mathrm{flight}.$

Maximum height (Hmax): It is the greatest height to which a projectile rise above the point of the projection.

At maximum height the vertical velocity of the projectile become zero i.e. Vy = 0 so,
${\mathrm{V}}^2\mathrm{y}={\mathrm{U}}^2\mathrm{y}–2\mathrm{gHmax}$
$\mathrm{or}, 0=(\mathrm{u}sin\mathrm{\theta }{)}^2–2\mathrm{gHmax}$
$\therefore \mathrm{Hmax}=\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}$ This is expression for maximum height.

Horizantal Range (R): The horizantal distance covred by the projectile during its time of flight is called horizantal range.

Since these is no accleration due to gravity in horizantal range or direction.
so,
R = Horizanal velocity X time of flight
$\mathrm{or}, \mathrm{R}=\mathrm{Ux}.\mathrm{T}$
$\mathrm{or}, \mathrm{R}=\mathrm{U}cos\mathrm{\theta }.\frac{2\mathrm{u}sin\mathrm{\theta }}{\mathrm{g}}$
$\therefore \mathrm{R} =\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}$ this is a expression for horizantal range.

Maximum Horizantal range (Rmax):

$\mathrm{If}, \mathrm{Sin}2\mathrm{\theta }=1$
$\mathrm{or}, \mathrm{Sin}2\mathrm{\theta }=\mathrm{Sin}90°$
$\Rightarrow \mathrm{\theta }=45°$
Thus, the horizantal range will be maximum if the projectile is fired at angle of $45°$with horizantal.

Two angle of projectile for the same horizamtal range:

We know that for the angle of projection $\mathrm{\theta }$ the horizantal range is given by:
$\mathrm{R}=\frac{{\mathrm{u}}^2\sin \mathrm{\theta }}{\mathrm{g}}$
Let $\mathrm{R}‘$be the horizantal range when the angle of projection is $90°–\mathrm{\theta }$ then,
$\mathrm{R}‘=\frac{{\mathrm{u}}^2\sin 2(90°–\mathrm{\theta })}{\mathrm{g}}$
$\mathrm{R}‘=\frac{{\mathrm{u}}^2\sin \left(180°–2\mathrm{\theta }\right)}{\mathrm{g}}$
$\mathrm{R}‘=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}$
$\therefore \mathrm{R}‘=\mathrm{R}$
Thus, we can see that for two angle of projection that is $\mathrm{\theta }$ and $90°–\mathrm{\theta }$ the horizantal range are same.

Velocity of any instant:

Horizantal velocity at any instant,
$\mathrm{Vx}=\mathrm{Ux}=\mathrm{U}cos\mathrm{\theta }$
vertical velocity at any instant,
$\mathrm{Vy}=\mathrm{Uy}–\mathrm{gt}$
$= \mathrm{U}sin\mathrm{\theta }–\mathrm{gt}$
$\therefore \mathrm{velocity} \mathrm{at} \mathrm{any} \mathrm{ins}tan\mathrm{t} \mathrm{V}=\sqrt{{\mathrm{V}}^2\mathrm{x}+{\mathrm{V}}^{2}y}$
If $\theta$ be the angle made by velocity with horizantal velocity, then,
$\mathrm{Tan}\mathrm{\theta }=\frac{\mathrm{Vy}}{\mathrm{vx}}$
$\therefore \theta =Ta{n}^{–1}\left(\frac{Vy}{Vx}\right)$

Numericals

Q.2[A] A projectile fired at angle 18° has certain horizantal range. State another angle of projection for the same horizantal range?

solution,

There are two angle of projection for the horizantal range.
When the angle of projection is ($\theta =180°$)
then,
$R=\frac{u^2\sin 2.18°}{g}=\frac{u^2\sin 36°}{g}$
Now,
When angle of projection is $(90°–\theta )=\left(90°–18°\right)=72°$
then,
$R‘=\frac{u^2\sin 2.(90°–18°)}{g}=\frac{u^2\sin 2.\left(180°–36°\right)}{g}=\frac{u^2\sin 36°}{g}$
Hence, another angle of projection is $72°$ for the same of horizantal range.

Q.2[B] What to be the effect on Rmax in doubling the intial velocity of a projectile?

We have,
$Rmax=\frac{u^2\sin 2.(45°)}{g}=\frac{u^2}{g}$
If ‘u’ or intial velocity is double u’=2u
or,$R‘max=\frac{(u{)}^2}{g}=\frac{(2u{)}^2}{g}=4\left(\frac{u^2}{g}\right)$
so, when doubling the inital velocity then Rmax is increase by 4 times with inital velocity .

Q.2[C] Find the angle of projection at which the horizantal range and Hmax of projection are equal?

solution,
Given
R = Hmax
$or, \frac{u^2\sin 2\theta }{g}=\frac{u^2{\sin }^2\theta }{2g}$
$or, Sin2\theta =\frac{{\sin }^2\theta }{2g}$
$or, \frac{2sin\theta .cos\theta }{\sin \theta }=\frac{\sin \theta }{2}$
$or, 4cos\theta =sin\theta$
$or, \frac{sin\theta }{cos\theta }=4$
$or, tan\theta =4$
$or, \theta =ta{n}^{–1}\left(4\right)$
$\therefore \theta =75.96°$
Hence, If R and Hmax is equal then angle of projection on is $75.96°$

Q.2[D] A batter hits a baseball so that it leaves the bat with an inital speed 37m/s at angle of 53°. Find the position OF ball and direction of its velocity after 2sec.Treat the base ball as projectile?

solution,
inital velocity(u)=37m/s
angle of projection ($\theta$)=$53°$
time(t)=2sec
now,
we have
x= Ux.t = Ucos$\theta$.t = 37 xcos$53°.2$=44.53m
and
$y=tan\theta x–\frac{g}{2u^2{\cos }^2\theta }.x^2$
$= tan53°\left(44.53\right)–\frac{10}{2\times (37{)}^2\times (cos53°{)}^2}\times (44.53{)}^2$
$= 59.09–19.99$
$= 39.1m$
Thus, position of the ball is an point p(44.53,39.1)
for the direction:
$Vx=Ux=Ucos\theta =37\times cos53°=22.26 and$
$Vy=Usin\theta –gt$
$=37\times Sin53°–10\left(2\right)$
$=9.549$
Now,
$\mathrm{\theta }={\mathrm{Tan}}^{–1}\left[\frac{\mathrm{Vy}}{\mathrm{Vx}}\right]=ta{n}^{–1}\left(\frac{9.549}{22.26}\right)=23.21°$
$\mathrm{so}, \mathrm{direction} \mathrm{is} \mathrm{\theta }=23.21°$

Q.2[E] A base ball is thrown towards a player with an inital velocity 20m/s and 45° with the horizantal at the moment the ball is thrown, the player is 50m from the thrower. At what speed and direction must he run to catch the ball at the same height at which it was released?

Given;

Inital velocity(u)= 20m/s
angle of projection $\left(\theta \right)=45°$
Now,
$R=\frac{u^2\sin 2\theta }{g}$
$R=\frac{(20{)}^2\times sin2(45°)}{10}$
$R=40m$
$Similarly;$
$T=\frac{2usin\theta }{g}$
$T=\frac{2\times 20\times sin45°}{10}$
$T=4\times \frac{1}{\sqrt{2}}$
$T=2\sqrt{2}\sec$
$Again,$
$Speed=\frac{distance}{time}=\frac{(50–40)m}{2\sqrt{2}s}=\frac{5}{\sqrt{2}}=3.53m/s$
$\therefore speed=3.53m{s}^{–1}$
Hence, speed player must run towards first player with speed 3.53m/s.

Q.2[E] A projectile is fired from the ground level with a velocity of 500m/s at 30° to the horizantal. Find the horizantal range and greatest vertical height to which it rise. What is the least speed with which it could be projected in order to acheive the same horizantal range?(g=10m/s)

given,

inital velocity (u)=500m/s
angle of projection $\left(\theta \right)=30°$
g=10m/s
Now,we have,
$R=\frac{u^2\sin 2\theta }{g}=\frac{(500{)}^2\times sin2(30°)}{10}$
$\therefore R=21651m$
similary;
$Hmax=\frac{u^2{\sin }^2\theta }{2g}=\frac{(500{)}^2\times si{n}^{2}30°}{2\times 10}$
$\therefore Hmax=3125m$
Again, the case of least speed;
by using formula
$R=\frac{u^2\sin 2\theta }{g},we get$[Here’u’ willbe least when $\sin 2\theta will be maximum the maximum value of sin2\theta =1$]
$R=\frac{{u^2}_{least}\sin 2\theta }{g}=\frac{{u^2}_{least}.1}{g}$
$or,{u^2}_{least}=Rg$
$or,u_{least}=\sqrt{2g}=\sqrt{21651\times 10}=\sqrt{216510}$
$\therefore u_{least}=465.30m/s$
Thus, least speed is 465.30m/s in same horizantal range.

Horizantal Projectile:

Any body thrown horizantally from the top of the tower and which moves only under the action of gravity is called horizantal projectile.

Let us consider a body is thrown horizantally from the top of tower of height’h’ and with velocity’u’. Suppose after time ‘t’ it reach at point p(x,y) at depth ‘y’ and horizantal distance x from point of projection.

Since it is thrwon horizantally so it’s inital vertical velocity u_y=0 and accleration due to gravity acts in vertical direction only so, it’s horizantal velocity remain constant.

#Motion along horizantal;
$or,x=u_{x}t$
$or,x=ut$
$or,t=\frac{x}{u}$…………..(i)
#Motion along vertical;
$or,y=–u_y+\frac{1}{2}g{t}^2$
$or,y=\frac{1}{2}g{t}^2.......\left(i\right)$
using equation (i) in equation (ii), we get,
$y=\frac{1}{2}g(\frac{x}{u}{)}^2$
$or,y=\frac{g}{2u^2}.x^2..........\left(iii\right)$
Which is the equation of the form $or,y=\frac{g}{2u^2}.x^2..........\left(iii\right)$ and it is equation of parabola.
so, path of horizantal projectile is also parabolic.

Time of Flight(T): The time for which projectile remains in space is called time for flight.

If ‘T’ be the time of fligt then,
$\mathrm{h}=–{\mathrm{u}}_{\mathrm{y}}\mathrm{T}+\frac{1}{2}{\mathrm{gT}}^2$
$\mathrm{h}=\frac{1}{2}{\mathrm{gT}}^2$
$\therefore T=\sqrt{\frac{2h}{g}} This is expression for time of flight.$

Horizantal Range (R);

The horizantal distance coverd by the projectile during it’s time of flight is called horizantal range.

Now,
R= Horizantal velocity X tiime of flight
$R=u_x.T$
$\therefore R=u\sqrt{\frac{2h}{g}}This is expression for horizantal range.$

Special Cases (I and II):

Fig: Case I
Time of Filght(T):
$h=–u_{y}T+\frac{1}{2}g{T}^2$
$h=–usin\theta .T+\frac{1}{2}g{T}^2$
Horizantal Range(R):
$R=u_{x}T$
$R=ucos\theta T$

Fig: caseII
Time of Flight(T):
$or,h=u_{y}T+\frac{1}{2}g{T}^2$
$or,h=usin\theta T+\frac{1}{2}g{T}^2$
Horizantal Range(R):
$R=u_{x}T$
$R=ucos\theta T$

Velocity of any instant:

Horizantal velocity at any instant ;
$\therefore {\mathrm{v}}_{\mathrm{x}}={\mathrm{u}}_{\mathrm{x}}=\mathrm{u}$
Vertical velocity at any instant;
${\mathrm{v}}_{\mathrm{y}}={\mathrm{u}}_{\mathrm{y}}+\mathrm{gt}$
$\therefore {\mathrm{v}}_{\mathrm{y}}=\mathrm{gt}$
$\therefore \mathrm{Velocity} \mathrm{at} \mathrm{any} \mathrm{ins}tan\mathrm{t}\mathrm{;}$
$\therefore \mathrm{v}=\sqrt{{{\mathrm{v}}^2}_{\mathrm{x}+}{{\mathrm{v}}^2}_{\mathrm{y}}}=\sqrt{{\mathrm{u}}^2+{\mathrm{g}}^2{\mathrm{t}}^2}$
$\mathrm{If} \mathrm{\theta } \mathrm{be} \mathrm{the} \mathrm{angle} \mathrm{made} \mathrm{by} \mathrm{velocity} \mathrm{with} \mathrm{horizantal}:\mathrm{then},$
$\tan \mathrm{\theta }=\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}$
$\mathrm{or},tan\mathrm{\theta }=\left(\frac{\mathrm{gt}}{\mathrm{u}}\right)$
$\therefore \mathrm{\theta }=ta{n}^{–1}\left(\frac{\mathrm{gt}}{\mathrm{u}}\right)$

Numericals:

Q.3[A] An aeroplane diving at an angle of 37° with the horizantal drops a mail bag at an height of 730m. The projectile hits the ground 5sec after being relased. What is the speed of air Craft?

we have;
$\mathrm{h}={\mathrm{u}}_{\mathrm{y}}\mathrm{T}+\frac{1}{2}{\mathrm{gT}}^2$
$\mathrm{or},730=\mathrm{u}.sin37°.5+\frac{1}{2}\times 10\times 5^2$
$\mathrm{or},730–125=5\mathrm{u}.sin37°$
$\mathrm{or},\frac{605}{5}=\mathrm{u}.sin37°$
$\mathrm{or},\mathrm{u}=\frac{121}{sin37°}$
$\therefore \mathrm{u}=201.05\mathrm{m}/\mathrm{s}$
hence, speed of the air craft is 201.05m/s

Q.3[B] An air plane is flying with a velocity of 90m/s at angle of 23° above the horizantal. When the plane is 114m directly above or dog that is standing on level ground. A sout-case drops out of lodge compartment. How far from the dog will the sout-case land. You can ignore air resistance.

height(h)= 114m
inital velocity(u)= 90m/s
angle of projection $\left(\mathrm{\theta }\right)=23°$
${\mathrm{u}}_{\mathrm{x}}=\mathrm{u}cos\mathrm{\theta }$
${\mathrm{u}}_{\mathrm{y}}=\mathrm{u}sin\mathrm{\theta }$
$–{\mathrm{u}}_{\mathrm{y}}=–\mathrm{u}sin\mathrm{\theta }$
we have
$\mathrm{h}=–{\mathrm{u}}_{\mathrm{y}}\mathrm{T}+\frac{1}{2}\times \mathrm{g}\times {\mathrm{T}}^2$
$\mathrm{or},114=–90sin23°\mathrm{T}+\frac{1}{2}10\times {\mathrm{T}}^2$
$\mathrm{or},5{\mathrm{T}}^2–35.16\mathrm{T}–114=0........\left(\mathrm{i}\right)$
$\mathrm{Comparing} \mathrm{equation} \left(\mathrm{i}\right) \mathrm{with} {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$
$\therefore \mathrm{a}=5, \mathrm{b}=–35.16 \mathrm{\&} \mathrm{c}=–114$
$\mathrm{so} \mathrm{u}\mathrm{sing} \mathrm{formula},$
$\mathrm{T}=\frac{–\mathrm{b}\pm \sqrt{{\mathrm{b}}^2–4\mathrm{ac}}}{2\mathrm{a}}$
$\mathrm{or},\mathrm{T}=\frac{35.16\pm \sqrt{(–35.16{)}^2–4(5\left)\right(–114)}}{2\times 5}$
$\mathrm{or},\mathrm{T}=\frac{35.16\pm \sqrt{3516.63}}{10}$
$\mathrm{or},\mathrm{T}=\frac{35.16\pm 59.3}{10}$
$\mathrm{taking} +\mathrm{ve},$
$\mathrm{T}=\frac{35.16+59.3}{10}$
$\therefore \mathrm{T}=9.4sec$
$\mathrm{putting} \mathrm{T}=9.4sec$
$\mathrm{R}=\mathrm{u}\times \mathrm{T}$
$\mathrm{R}=\mathrm{u}cos\mathrm{\theta }\mathrm{T}$
$\mathrm{R}=90.cos23°.9.4$
$\therefore \mathrm{R}=782.55\mathrm{m}$
Hence, 782.55m far from the dog will the shout-case land in 9.4 sec time taken.

Q.3[C] A body is projected horizantally from the top of a tower 100m high with a velocity of 9.8m/s. Find the velocity with which it’s hits the ground?

solution,
Given
h=100m
inital velocity$\left({\mathrm{u}}_{\mathrm{x}}\right)=\mathrm{u}=9.8\mathrm{m}/\mathrm{s}$
Now,
$\mathrm{T}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}=\sqrt{\frac{2\times 100}{10}}=\sqrt{20}$
$\therefore \mathrm{T}=4.47sec$
We know in the case of velocity at any instat;
$\mathrm{v}=\sqrt{{{\mathrm{v}}^2}_{\mathrm{x}}+{{\mathrm{v}}^2}_{\mathrm{y}}}$
$\mathrm{v}=\sqrt{{\mathrm{u}}^2+(\mathrm{gt}{)}^2}$
$\mathrm{v}=\sqrt{(9.8{)}^2+(10\times 4.47{)}^2}$
$\mathrm{v}=\sqrt{2094.13}$
$\therefore \mathrm{velocity}=45.76\mathrm{m}/\mathrm{s}$
$\mathrm{moreover}\mathrm{;}$
$\mathrm{or},\tan \mathrm{\theta }=\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}$
$\mathrm{or},\tan \mathrm{\theta }=\frac{\mathrm{gt}}{\mathrm{u}}$
$\mathrm{or},\mathrm{\theta }=ta{n}^{–1}\left[\frac{10\times 4.47}{9.8}\right]$
$\mathrm{or},\mathrm{\theta }=ta{n}^{–1}\left(4.56\right)$
$\therefore \theta =77.63°$
Hence, Velocity at this instant is 45.76and direction$77.63°$

Relative Velocity :

The velocity of one body related to another body is called relative velocity.

Let us consider two bodies A & B moving with velocities $\overrightarrow{V_A} \&\overrightarrow{V_B}$ inclined at an angle $\theta$. Now the velocity of body A related to body Bis denoted by$\overrightarrow{V_{AB}}$ and given by:
$\overrightarrow{V_{AB}}=\overrightarrow{V_A} –\overrightarrow{V_B}$
similary,
Velocity of body related to body A is denoted by $\overrightarrow{V_{BA}}$ & defined as
$\overrightarrow{V_{BA}}=\overrightarrow{V_B} –\overrightarrow{V_A}$

To find $\overrightarrow{V_{AB}}$

Let us consider reverse $\overrightarrow{V_B}$ and complete the parallelogram then diagonal Represent $\overrightarrow{V_{AB}}$
Now,
using paralleogram law of vector addition;
$\overrightarrow{V_{AB}}=\sqrt{{V^2}_A+{V^2}_B+2 V_A{V}_B.cos(180°–\theta )}$
$\overrightarrow{V_{AB}}=\sqrt{{V^2}_A+{V^2}_B–2 V_A{V}_{B }\cos \theta }$
Case I: When two bodies are in same direction i.e,$(\mathrm{\theta }=0°)$
$\overrightarrow{V_{AB}}=\sqrt{{V^2}_A+{V^2}_B–2 V_A{V}_B.cos0°}$
$\overrightarrow{V_{AB}}=\sqrt{{V^2}_A+{V^2}_B–2 V_A{V}_{B }}$
$\overrightarrow{V_{AB}}=\sqrt{(V_A–V_B{)}^2}$
$\therefore \overrightarrow{V_{AB}}=V_A–V_B$
Hence, when two bodies are in same direction then resulatant is $V_A–V_B$ or decrease .
Case II: When two bodies are in opposite direction; i.e($\mathrm{\theta }=180°$)
$\overrightarrow{V_{AB}}=\sqrt{{V^2}_A+{V^2}_B–2 V_A{V}_B.cos180°}$
$\overrightarrow{V_{AB}}=\sqrt{{V^2}_A+{V^2}_B+2 V_A{V}_{B }}$
$\overrightarrow{V_{AB}}=\sqrt{(V_A+V_B{)}^2}$
$\therefore \overrightarrow{V_{AB}}=V_A+V_B$
Hence, when two bodies are in opposite direction then resultant or relative velocity is $V_A+V_B$ or increase.