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

projectile_1_1

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

projectile_2_1There 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

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s