Projectile motion (Motion on 2D)

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

$h=u_y.t-\frac{1}{2}g{t}^2$

or, $0=t(u_y-\frac{1}{2}gt)$

or, $t=\frac{2u_y}{g}=\frac{2usin\theta }{g}$

so time to reach maximum height = $\frac{usin\theta }{g}$

Maximum height

Maximum height during the journey of projectile.

At maximum height : vertical velocity=0, horizontal velocity=$ucos\theta$

Using equation of motion on 1D ( vertical direction)

$v^2=u^2-2gH$

$v=0,u=usin\theta ,h=H_{\text{max}}$

so, $H_{\text{max}}=\frac{(usin\theta {)}^2}{2g}=\frac{u^{2}si{n}^2\theta }{2g}..........(ii)$

Range of projectile

It is the horizontal distance travelled by the projectile.

Range(R)= Horizontal distance * time taken

or, $R=ucos\theta \ast \frac{2usin\theta }{g}$

or, $R=\frac{2u^{2}sin\theta \ast cos\theta }{g}$

or, $R=\frac{u^{2}sin2\theta }{g}$...........(iii)

For maximum range $sin2\theta =1$, $\theta =45°$

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

Equation of trajectory

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

$x=ucos\theta \ast t$

Then vertical distance corresponding to horizontal distance x is

$y=usin\theta \ast t-\frac{1}{2}g{t}^2=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$

which is on the form $y=ax-b{x}^2$, is parabolic.

where, $a=tan\theta$ and $b=\frac{g{x}^2}{2u^{2}co{s}^2\theta }$

$\therefore$Range(R)=$\frac{a}{b}=\frac{tan\theta }{\frac{g{x}^2}{2u^{2}co{s}^2\theta }}=\frac{u^{2}sin2\theta }{g}$

$\therefore$Maximum height=$\frac{a^2}{4b}=\frac{(tan\theta {)}^2}{4.\frac{g{x}^2}{2u^{2}co{s}^2\theta }}=\frac{u^{2}si{n}^2\theta }{2g}$

Dividing equation (ii) and (iii)

$\frac{H}{R}=\frac{tan\theta }{4}$

Velocity after certain time t is:

Horizontal velocity= constant=$ucos\theta =v_x$

Vertical velocity= $usin\theta -2g=v_y$

so, the net velocity v after time t is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(ucos\theta {)}^2+(usin\theta -2g{)}^2}$

and the angel made by the velocity is $tan\alpha =\frac{v_y}{v_x}=\frac{usin\theta -2g}{ucos\theta }$

Energy during projectile motion

Initially, Total energy=Kinetic energy=$\frac{1}{2}m{u}^2$

During motion at highest point

K.E.=$\frac{1}{2}m(ucos\theta {)}^2$=$\frac{1}{2}m{u}^{2}co{s}^2\theta$

P.E.=mgh=mg$\frac{u^{2}si{n}^2\theta }{2g}$=$\frac{1}{2}m{u}^{2}si{n}^2\theta$

T.E. = K.E. + P.E. =$\frac{1}{2}m{u}^{2}co{s}^2\theta +\frac{1}{2}m{u}^{2}si{n}^2\theta$=$\frac{1}{2}m{u}^2$

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;

$y=v_y\ast t+\frac{1}{2}g{t}^2$

or, $y=\frac{1}{2}g\frac{x^2}{u^2}=k{x}^2$

which is parabolic in nature.

At point p, $v_x=u$,$v_y=0+gt$

so, resultant velocity v=$\sqrt{v_x^2+v_y^2}=\sqrt{(u{)}^2+(gt{)}^2}$

And angle made by velocity

$tan\alpha =\frac{v_y}{v_x}=\frac{gt}{u}$

Time of flight = $\sqrt{\frac{2H}{g}}$

Range=u.t=u*$\sqrt{\frac{2H}{g}}$