Class XI Notes on Projectile Motion

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Kinematics vectors force 10inertia Projectile Motion momentummqwertyuiopasdfghjklzxlmqwertyuiopaprojectile motion one dimentionl 4/10/2012 motion circular motion banking of roads By Salik Hussain vectors force inertia momentumprojectile motion one dimentionl copyrighted 2012© motion circular motion banking of roads vectors force inertia momentumprojectile motion one dimentionl motion circular motion banking of roads vectors force inertia momentum Ayaan Hassan dfghjk Physicsmqwertprojectile motion one 10Qwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnproject  ile motion one dimentionl motion circular motion banking ba nking of roads dimentionl motion circular motion banking of roads vectors force inertia10 momentum Ayaan Hassan dfghjk  Physicsmqwertprojectile motion one dimentionl motion circular motion banking of roads vectors force inertia momentum Ayaan Hassan dfghjk Physicsmqwertprojectile motion one Qwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnprojectile motion one dimentionl motion circular motion banking of roads Qwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnprojectile motion one dimentionl motion circular motion banking of roads dimentionl motion circular motion banking of roads vectors force inertia momentum Ayaan aazaad Hassan dfghjk  Physicsmqwertprojectile motion one wert uio asdf h klzxcvbnm wert uio asdf h klzxcvbn ro ectile

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 Projectile Motion 

By Salik Hussain

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: A packet released from an airplane in flight. A golf ball in flight. A bullet fired from a rifle. A jet of water from a hole near the bottom of a water tank. A bomb dropped from a fighter plane in flight Projectile motion is a case of two-dimensional two-dimensional motion. Any case of  two-dimensional 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, effect of rotation of projectile, effect of curvature of earth, effect of  rotation of earth. We will take x-axis along the horizontal direction and y-axis along the vertical direction.

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 Projectile Motion 

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By Salik Hussain

Case 1: Horizontlal 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.

Figure showing the case of Horizontal Motion Motion.”A Free Body Diagram” is also shown

Equation of a trajectory(path of a projectile) for Horizontal Projection: We know at any instant , x = ut

» t=x/u 2

Also, y= ( 1/2)gt

Subsituting for ‘t’ we get 







2

y= ( )g( ) 

2

2

» y= ( )(g/u )x 

2

»y= kx 2 ;where k= g/(2u ) 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.

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 Projectile Motion 

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By Salik Hussain

Table showing the motion along both axes:

Along x-axis

Along y-axis 

1. Component of initial velocity along x-axis.

1. Component of initial velocity along y-axis.

Ux= u

Uy=0

2. Acceleration along x-axis ax=0 (Because no force is acting along the horizontal direction) 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. 4. The displacement along x-axis at any instant t 2 x=uxt + (1/2) axt » x=uxt + 0 » x=uxt

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2. Acceleration Acceleration along y-axis 2 ay=g=9.8m/s It is directed downwards. 3. Component of velocity along the y-axis at any instant t. Vy=uy + ayt=0 + gt » vy=gt

4. The displacement along y-axis at any instant t 2 y= uyt + (1/2) ay t 2 » y= 0 + (1/2) ayt 2 » y=1/2 gt

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 Projectile Motion 

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By Salik Hussain

Figure showing components components of velocity of a Horizontally fired fired Projectile:

Net velocity at any instant of time ,’t’: We know, at any instant, t ’

’

vx= u vy= gt » v= (vx2 + vy2)1/2 = [u2 + (gt)2] 1/2 Direction Direction of v with the horizontal at any instant :

-1

-1

Ø = tan (vy/vx)= tan (gt/u)

Time of flight(Tf): 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. 2

From y= uyt + (1/2) gt When , y= h , t=Tf 

[y=h, means that the projectile has hit the ground. How?]

2

» h= 0 + (1/2) gt 1/2

» T=(2h/g) 5|Page

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 Projectile Motion 

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By Salik Hussain

Range(R): It is the horizontal distance covered during the time of flight T. From x= ut When t=Tf  , x=R [Tf is the Time of flight] »Range, R=uTf  1/2

» R=u(2h/g)

[because Tf =(2h/g)

1/2

]

Case 2: Angular Projection Projection

We will now 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 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. 6|Page

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 Projectile Motion  Along X-axis

By Salik Hussain

Along Y-axis 

1. Component of initial velocity along x-axis. Ux= u cosØ

1. Component of initial velocity along y-axis. Uy=u sinØ

2. Acceleration along x-axis ax=0 (Because no force is acting along the horizontal  direction) 3. Component of velocity along the x-axis at any instant t. Vx=ux + axt =ucosØ + 0= ucosØ » vx=ucosØ This means that the horizontal component of  velocity does not change throughout the projectile motion. 4. The displacement along xaxis at any instant t 2 x=uxt + (1/2) axt » x=ucosØ.t

2. Acceleration along y-axis 2 ay= -g= -9.8m/s (g is negative as it is acting in the downward direction)

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3. Component of velocity along the y-axis at any instant t. Vy=uy + ayt » vy=usinØ - gt

4. The displacement along yaxis at any instant t 2 y= uyt + (1/2) ayt 2 » y= usinØ.t - (1/2)gt

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 Projectile Motion 

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By Salik Hussain

Equation of Trajectory (Path of projectile) for Angular Projection: At any instant t, x= ucosØ.t » t= x/(ucosØ) Also , y= usinØ.t - (1/2)gt

2

Substituting for t, 2

y= usinØ.x/(ucosØ) -- (1/2)g[x/(ucosØ)] 2

2

2

» y= x.tanØ -- [(1/2)g.sec Ø.x ]/u

2

This equation is of the form y= ax + bx 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’: vx=ucosØ vy=usinØ – gt

» V= (vx2 + vx2 )1/2 Direction of v with the horizontal at any instant : ß= tan

-1

(vy/vx) ,where ‘ß’ is the angle that the resultant velocity(v)

makes with the horizontal at any instant .

Time of flight,Tf: Angular Projectile motion is symmetrical about the highest point.The object will reach the highest point in time Tf /2 .At the highest point,the vertical component of velocity vy

becomes equal to zero. vy =usinØ - gt 8|Page

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 Projectile Motion 

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By Salik Hussain

At t=T f /2 , vy= 0 0= usinØ – gTf /2 » T= (2usinØ)/g

[ALTERNATE APPROACH: Suppose the particle is at B, as shown in figure below. The equation for Horizontal motion gives, OB=x=Uxt=ucosθ They y-coordinate at the point B is zero. Thus, from the t he equation of  vertical motion,

2

y=utsinθ-1/2gt

or, 0=utsinθ-1/2gt

2

or, t(usinθ-1/2gt)=0 Thus, either t=0 or t= 2usinθ/g

Now t=0, corresponds corresponds to the position O of the particle. The time at which it reaches B is thus, Tf =2usinθ/g, which is same as by other method!]

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 Projectile Motion 

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By Salik Hussain

Maximum height,H max: Equation for vertical distance (y-component), 2

y= uyt - (1/2)gt

At t=Tf /2 , y=H 2

» Hmax.= usinØ.Tf /2 - (1/2)g(Tf /2) Substituting Tf 

2

Hmax.= usinØ.usinØ/g - (1/2)g(usinØ/g)

[Since Tf =2 usinØ/g, usinØ/g, so T f /2=usinØ/g ]

= (u2sin2Ø)/g - (u2sin2Ø)/2g 2

2

» Hmax.= (u sin Ø)/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=Tf , x=R R= uxTf  = ucosØ.2usinØ/g

[ux=cosØ & Tf = 2usinØ/g]

2

= 2u sinØcosØ/g 2

=u (2 sinØcosØ)/g 2

= u sin2Ø/g

[using 2sinØcosØ= sin2Ø]

2

» R= (u sin2Ø)/g

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