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Projectile motion examples and formulas

What is projectile motion?

“Projectile motion is two dimensional motion under constant acceleration.”or When a body is thrown vertically upward then it follows a curved path such a motion of body is called projectile motion. Up till now we have been studying the motion of a particle along a straight line i.e motion in one dimension.Now we consider the motion of the ball,when it is thrown horizontally from certain height.It is observed that the ball travels forward as well as falls downwards,until it strikes something.projectile motionSuppose that the ball leaves the hand of the thrower at point A,and that its velocity at that instant is completely horizontal.Let this velocity be Vx .According to Newton’s first law of motion,there will be no acceleration in horizontal direction,unless a horizontally directed force acts on the ball.Ignoring the air friction,only force acting on the ball during flight is the force of gravity.There is no horizontal force acting on it.So its horizontal velocity will remain unchanged and will be Vx,until the ball hits something.The horizontal motion of ball is simple.The ball moves with constant horizontal velocity component.Hence horizontal distance X is given by:

X = V t

The vertical motion of the ball is also not complicated.It will accelerate downward under the force of gravity and hence a = g .This vertical motion is the same as for a freely falling body.Since initial velocity is zero,hence ,vertical distance y by using second equation of motion is given by:

formula of projectile height

It is not necessary that an object should be thrown with some initial velocity in the horizontal direction.A football kicked off by a player;a ball thrown by a cricketer and a missile fired from a launching pad,all projected at some angles with the horizontal,are called projectiles.

In such cases ,the motion of a projectile can be studied easily by resolving it into horizontal and vertical components which are independent of each other.Suppose that a projectile is fired in a direction angle θ with the horizontal by velocity V along the horizontal and vertical components of velocity V along the horizontal and vertical directions be V V cosθ and V sinθ respectively.projectile motion 2The horizontal acceleration is ax = 0 because we have neglected air resistance and no other force is acting along the direction resistance and no other force is acting along this direction whereas vertical acceleration ay = g.Hence,the horizontal component Vx remains constant and at any time t,we have:

Vfx = Vix = Vicosθ

Now we consider the  vertical motion.The initial vertical component of the velocity is V sinθ in the upward direction.By using first equation of motion (Vf = Vi + at) the vertical component Vfy of the velocity at any instant t is given by:

Vfy = Vsinθ  – gt

The magnitude of velocity at any instant is:

V =Vfx2 + Vfv2

The angle Φ which this resultant velocity makes with the horizontal can be found from:

projectile direction formula

In projectile motion one may wish to determine the height to which the projectile rises ,the time of flight and horizontal range.

Height of  the projectile:

In order to determine the maximum height of the projectile attains,we use the equation of motion:

2 as =Vf –  Vi2

As body moves upward ,so a = -g ,the initial vertical velocity Viy =Vi sinθ  and Vfy =0 because the body comes to rest after reaching the highest point.Since:

S = height = h

-2gh= 0 -Vi 2sin2θ

h=Vsin2θ/2g

Time of flight of the projectile:

“It is the total time for which the projectile remains in air.”or The time taken by the body to cover the distance from the place of its projection to the place where it hits the ground at the same level is called the time of flight.

This can be obtained by taking S =h  =0,because the body goes up and comes back to same level,thus covering no vertical distance.If the body is projecting with velocity V making angle θ with a horizontal ,then its vertical component will be Visinθ.Hence the equation is:

time of flight

Range of the projectile:

Maximum distance which a projectile covers in the horizontal direction is called the range of the projectile.Or The horizontal distance covered by the projectile between point of projection and point of return to level of projection is called “Range” of the projectile and is represented by “R”.

To determine the range R of the projectile,we multiply the horizontal component of the velocity of projection with total time taken by the body after leaving the point of projection.Thus:

X={ V }_{ ix }t\quad +\frac { 1 }{ 2 } { a }_{ x }{ t }^{ 2 }\quad .............(1)\\ X=R\\ { V }_{ ix }={ V }_{ 0 }Cos\theta \\ we\quad know\quad that:\\ t=T=\frac { 2{ V }_{ 0 }Sin\theta  }{ g } \\ { a }_{ x }=0\\ put\quad vlues\quad of\quad { V }_{ ix }\quad ,t\quad and\quad { a }_{ x }\quad in\quad (1)\quad we\quad get:\\ R={ V }_{ 0 }Cos\theta \quad \times \quad \frac { 2{ V }_{ 0 }Sin\theta  }{ g } +\frac { 1 }{ 2 } \times 0\times { (\frac { 2{ V }_{ 0 }Sin\theta  }{ g } ) }^{ 2 }\\ R=\frac { { V }_{ 0 }^{ 2 } }{ g } (2sin\theta Cos\theta )+0\\ we\quad know\quad that\quad sin2\theta =2sin\theta cos\theta \\ R=\frac { { V }_{ 0 }^{ 2 } }{ g } Sin2\theta \quad \quad ..............(2)

Thus the range of the projectile depends upon the velocity of projection and the angle of projection.

Therefore:

R=\frac { { V }_{ 0 }^{ 2 } }{ g } Sin2\theta \quad \quad

For the range R to be maximum ,the factor sin2θ should have maximum value which is 1 when 2θ =90° or θ=45º

Watch also video about projectile motion:

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