# Projectile Motion

• ### Introduction

Projectile motion is a 2D motion. Common examples are football hit by a player, bullet fired from a gun, catapult, water coming out of hose pipe, mud particles thrown away from wheels of a vehicle, a high-jump athlete etc.

Note that aeroplane/spacecraft are NOT projectiles, because the propulsive force can be adjusted after their launch. In all previous cases, once the object is launched, its motion is purely under gravitational influence.

• ### Analysis

Let the velocity of projection be $\vec V$ and the angle of projection be $\theta$. The motion can be resolved into 2 motions:

a) along horizontal direction : This is a uniform velocity motion, $V cos \ \theta$

b) along vertical direction : This is a uniformly accelerated motion, with initial velocity $V sin \ \theta$ and acceleration $\vec g$

This resolution makes the analysis simpler. We find the following factors:

1) Max. height reached $H$

At the top, vertical velocity $=0$. Using $v^2 = u^2 + 2 as$,

${H = \frac {V^2 sin^2 \ \theta}{2g}}$

2) Time of flight $T$

Time required to reach the top will be $T/2$. Using $v = u + at$,

${\frac T 2 = \frac {V sin \ \theta}{g}}$

So,

${T = \frac {2 V sin \ \theta}{g}}$

3) Horizontal range $R$

This is the distance between the point of launch and the point where the projectile lands. Using $s = ut$,

${R = V cos \ \theta \times \frac {2V sin \ \theta}{g} = \frac {V^2 sin \ 2 \theta} {g}}$

The maximum values of $R$ and $H$ can be obtained by substituting maximum values of trigonometric functions, $sin \ \theta$ and $cos \ \theta$.

4) Equation of trajectory

Using $s = ut + \frac 1 2 a t^2$,

${x = V cos \theta \ t \ and \ y = V sin \theta \ t - \frac 1 2 gt^2}$

On eliminating $t$, we get,

$y = x \ tan \theta - \frac {g x^2}{2 V^2 cos^2 \theta}$

This is of the form $y = Ax - Bx^2$, which is a parabola.

• ### Analysis of motion, when the ground is not horizontal, but inclined at an angle $\phi$ with the horizontal

There are 2 ways of analyzing this motion:

I) Resolving the acceleration vector $\vec g$ along the ground and perpendicular to the ground :

In this case, both motions are uniformly accelerated, $\vec a_x = g \ cos \ \phi$ and $\vec a_y = g \ sin \ \phi$

II) Considering the equation of ground as $y = tan \ \phi \ x$

All quantities can be found by considering equations of parabola and straight line.

1) Time of flight, ${T = \frac {2 V sin \theta}{g cos \phi}}$

2) Range along plane, ${R = \frac {2 V^2 sin \theta}{g cos^2 \phi} \times cos (\theta + \phi)}$

For maximum range, ${\theta + \frac \phi 2 = \frac {\pi}{4} = 45^o}$

3) Max. height reached (perpendicular to ground),  ${H = \frac {V^2 sin^2 \theta}{2 g cos \phi}}$

• ## Few more important points

I) Throughout this analysis, air resistance, effect of wind and effect of rotation of earth are neglected.

II) For same range, there are 2 angles of projections possible. These angles are complementary, i.e. $\theta_1 + \theta_2 = \frac \pi 2$

III) At any point, the radius of curvature is ${\rho = \frac {V_{at \ that \ point}^2}{g cos \ \zeta}}$. $\zeta$ is the angle made by velocity at that point with horizontal.

IV) Minimum radius of curvature is at the max. height

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