Navigation

Mechanics

Heat and Thermodynamics

Magnetism

Geometrical Optics

Wave Optics

Electrostatics

Current Electricity

**Projectile motion (Motion on 2D) air resistance neglected**

When the velocity and acceleration of the body acts except $0°$and $180°$consists of projectile motion.

During projectile motion horizontal velocity and total energy remains same.

Two types of projection are considered in the chapter:

(i) Oblique projection

(ii) Horizontal projection from height

i. Oblique projection

**Time of projectile**

total vertical height=0

Using equation of motion on straight line

$h=u_{y}.t−21 gt_{2}$

or, $0=t(u_{y}−21 gt)$

or, $t=g2u_{y} =g2usinθ $

so time to reach maximum height = $gusinθ $

**Maximum height**

Maximum height during the journey of projectile.

At maximum height : vertical velocity=0, horizontal velocity=$ucosθ$

Using equation of motion on 1D ( vertical direction)

$v_{2}=u_{2}−2gH$

$v=0,u=usinθ,h=H_{max}$

so, $H_{max}=2g(usinθ)_{2} =2gu_{2}sin_{2}θ ..........(ii)$

**Range of projectile **

It is the horizontal distance travelled by the projectile.

Range(R)= Horizontal distance * time taken

or, $R=ucosθ∗g2usinθ $

or, $R=g2u_{2}sinθ∗cosθ $

or, $R=gu_{2}sin2θ $...........(iii)

For maximum range $sin2θ=1$, $θ=45°$

For angles $θ$and (90-$θ$) has the same range.

**Equation of trajectory**

Let x be the horizontal distance travelled at any time t, then,

$x=ucosθ∗t$

Then vertical distance corresponding to horizontal distance x is

$y=usinθ∗t−21 gt_{2}=xtanθ−2u_{2}cos_{2}θgx_{2} $

which is on the form $y=ax−bx_{2}$, is parabolic.

where, $a=tanθ$ and $b=2u_{2}cos_{2}θgx_{2} $

$∴$Range(R)=$ba =2ucosθgx tanθ =gu_{2}sin2θ $

$∴$Maximum height=$4ba_{2} =4.2ucosθgx (tanθ)_{2} =2gu_{2}sin_{2}θ $

Dividing equation (ii) and (iii)

$RH =4tanθ $

Velocity after certain time t is:

Horizontal velocity= constant=$ucosθ=v_{x}$

Vertical velocity= $usinθ−2g=v_{y}$

so, the net velocity v after time t is

$v=v_{x}+v_{y} =(ucosθ)_{2}+(usinθ−2g)_{2} $

and the angel made by the velocity is $tanα=v_{x}v_{y} =ucosθusinθ−2g $

**Energy during projectile motion**

Initially, Total energy=Kinetic energy=$21 mu_{2}$

During motion at highest point

K.E.=$21 m(ucosθ)_{2}$=$21 mu_{2}cos_{2}θ$

P.E.=mgh=mg$2gu_{2}sin_{2}θ $=$21 mu_{2}sin_{2}θ$

T.E. = K.E. + P.E. =$21 mu_{2}cos_{2}θ+21 mu_{2}sin_{2}θ$=$21 mu_{2}$

Hence total energy of projectile is conserved.

similarly, energy could be find out at other time.

**Horizontal projection from height**

At time t, horizontal distance travelled x=u*t

Then vertical distance corresponding to horizontal distance x is;

$y=v_{y}∗t+21 gt_{2}$

or, $y=21 gu_{2}x_{2} =kx_{2}$

which is parabolic in nature.

At point p, $v_{x}=u$,$v_{y}=0+gt$

so, resultant velocity v=$v_{x}+v_{y} =(u)_{2}+(gt)_{2} $

And angle made by velocity

$tanα=v_{x}v_{y} =ugt $

Time of flight = $g2H $

Range=u.t=u***$g2H $