Circular Motion Projectile Motion Work Energy Power

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Projectile

Projectile is a body thrown with an initial velocity in the vertical plane and then it moves in two dimensions under the action of gravity alone without being propelled by any engine or fuel. Its motion is called projectile motion. The path of a projectile is called its trajectory.

Examples:

1. A packet released from an airplane in flight.

2. A golf ball in flight.

3. A bullet fired from a rifle.

4. A jet of water from a hole near the bottom of a water tank.

Projectile motion is a case of two-dimensional motion .Any case of two dimensional motion can be resolved into two cases of one dimensional motion -one along the x-axis and the other along the y-axis.The two cases can be studied separately as two cases of one dimensional motion.The results from two cases can be combined using vector algebra to see the net result

What is important to remember is that the motion along the horizontal direction does not affect the motion along the vertical direction and vice versa.Horizontal motion and vertical motion are totally independent of each other .

A body can be projected in two ways :

1. Horizontal projection-When the body is given an initial velocity in the horizontal direction only.

2. Angular projection-When the body is thrown with an initial velocity at an angle to the horizontal direction.

We will study the two cases separately.We will neglect the effect of air resistance.We will take x-axis along the horizontal direction and y-axis along the vertical direction.

Case 1Horizontal Projection

A body is thrown with an initial velocity u along the horizontal direction.We will study the motion along x and y axis separately.We will take the starting point to be at the origin.

Along x-axis	Along y-axis
1. Component of initial velocity along x-axis. ux=u	1. Component of initial velocity along y-axis. uy=0
2. Acceleration along x-axis ax=0 (Because no force is acting along the horizontal direction)	2.Acceleration along y-axis ay=g=9.8m/s2 It is directed downwards.
3. Component of velocity along the x-axis at any instant t. vx=ux + axt =u + 0 vx=u This means that the horizontal component of velocity does not change throughout the projectile motion.	3. Component of velocity along the y-axis at any instant t. vy=uy + ayt =0 + gt vy=gt
4. The displacement along x-axis at any instant t x=uxt + (1/2) axt2 x=uxt + 0 x=u t	4. The displacement along y-axis at any instant t y= uyt + (1/2) ayt2 y= 0 + (1/2) ayt2 y=1/2gt2

Equation of a trajectory(path of a projectile)

We know at any instant x = ut
t=x/u

Also, y= (1/2)gt²

Subsituting for t we get

y= (1/2)g(x/u)²

y= (1/2)(g/u²)x²

 

y= kx² where k= g/(2u² )

This is the equation of a parabola which is symmetric about the y-axis.Thus,the path of projectile,projected horizontally from a height above the ground is a parabola.

 

Net velocity at any instant of time t

 

 

 

 

 

 

We know ,at any instant t

v_x= u
v_y= gt
v= (v_x² + v_y²)^1/2 = [u² + (gt)²]^1/2

 

^{Direction of v with the horizontal at any instant :}

(angle) = tan^-1 (v_y/v_x)= tan^-1 (gt/u)

Time of flight (T):
It is the total time for which the projectile is in flight ( from O to B in the diagram above)

To find T we will find the time for vertical fall
From y= u_yt + (1/2) gt^{2
When , y= h , t=T}

h= 0 + (1/2) gt²

T= (2h/g)^1/2

^{Range (R) :
It is the horizontal distance covered during the time of flight T.}

^{From x= ut
When t=T , x=R}

R=uT

R=u(2h/g)^1/2

^{Case 2 Angular Projection}

^{We will now consider the case when the object is projected with an initial velocity u at an angle to the horizontal direction.}

^{We assume that there is no air resistance .Also since the body first goes up and then comes down after reaching the highest point , we will use the Cartesian convention for signs of different physical quantities.The acceleration due to gravity 'g' will be negative as it acts downwards.}

^{We will separate the motion into horizontal motion (motion along x-axis) and vertical motion (motion along y-axis) .We will study x-motion and y-motion separately.}

 

Equation of Trajectory (Path of projectile)

At any instant t
x= ucosΦ.t
t= x/(ucosΦ)

Also , y= usinΦ.t - (1/2)gt²

Substituting for t
y= usinΦ.x/(ucosΦ) - (1/2)g[x/(ucosΦ)]²

y= x.tanΦ - [(1/2)g.sec².x² ]/u²

This equation is of the form y= ax + bx² where 'a' and 'b are constants.This is the equation of a parabola.Thus,the path of a projectile is a parabola .

Net velocity of the body at any instant of time t

v_x=ucosΦ
v_y=usinΦ - gt
v= (v_x² + v_y²)^1/2

Φ= tan^-1(v_y/v_x) Where Φ is the angle that the resultant velocity(v) makes with the horizontal at any instant .

Time of flight T

Angular Projectile motion is symmetrical about the highest point.The object will reach the highest point in time T/2 .At the highest point,the vertical component of velocity v_y becomes equal to zero .

v_y=usinΦ - gt
At t=T/2 , v_y= 0
0= usinΦ - gT/2

T= (2usinΦ)/g

Maximum height H

Equation for vertical distance (y component)
y= u_yt - (1/2)gt²

^At t=T/2 , y=H
H= usinΦ.T/2 - (1/2)g(T/2)^{2
substituting} T
H= usinΦ.usinΦ/g - (1/2)g(usinΦ/g)²
= (u²sin²)/g - (u²sin²)/2g

H= (u²sin²)/2g

Range R

Range is the total horizontal distance covered during the time of flight.

From equation for horizontal motion, x=u_xt
When t=T , x=R
R= u_xT = ucos.2usinΦ/g
= u²2sinΦcosΦ/g = u²sin2Φ/g using 2sinΦcosΦ= sin2Φ

R= (u²sin2Φ)/g