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# Projectile Motion Class 11 CBSE Physics Chapter 4 Motion in a Plane

Projectile motion is one of type of motion in a two-dimensional plane under an impact of gravitational force, it is the path of motion traversed by a thrown object towards the sky by an angle θ from the horizon. The path of the object is maintained by two components of the velocity, a vertical component towards the direction of the y-axis and a horizontal component towards the x-axis, both of these components are responsible to form a curved path of the object, such a motion is known as projectile motion.

## Projectile Motion Class 11 CBSE Physics Chapter 4 Motion in a Plane

The path of the trajectory of the projectile is given as follows

Let the angle of projection is θ, and the object is thrown with the u velocity, since the motion is in two-dimensional plane Y and X axis, therefore considering the two component of velocity one is towards the y-axis and another towards the x-axis.

Let the components of motion towards the y-axis is uy and  towards the x-axis is ux

uy  = usinθ (vertical component of velocity)and ux = ucosθ(horizontal component of velocity)

In this motion, acceleration is due to the gravity is along the y-axis,in x -direction its value is 0,ax=0

Since there is no acceleration in the x direction, therefore the velocity ux = ucosθ will remain constant and in the direction of y(i.e for the vertical component) the acceleration is always -g,ay = -g

The initial velocity of the obect in y direction is usinθ and when it obtained maximum height H(i.e point B),the vertical component of velocity,vy becomes 0,vy = 0

From A to B the velocity of the object decreases and from B to C ,the velocity of the object decreases

Note: The parameters are considered in this motion supposing air resistance is zero

Let the time taken by the flight of the object is =T

Maximum height (the distance of point B from the ground)=H

The range of the object(horizontal distance from A to C)=R

Let’s consider the motion from A to B.

The maximum height of the object is gained due to the y component of the velocity

The initial y component of the velocity is uy = usinθ and the final velocity (i.e at B) =Vy =0

Applying the first equation of the motion

v = u +at

0 = usinθ -gt

gt = usinθ

t = usinθ/g

The time taken from A to B and B to C will be the same due to the conservational law of energy

∴T = 2usinθ/g

Now applying third equation of the motion for determining maximum height of the projectile

v² = u² +2as

The initial y component of the velocity is uy = usinθ and the final velocity (i.e at B) =Vy =0

0= (usinθ)² +2(-g)H

2gH = u²sin²θ

H = u²sin²θ/2g

Since the range, AC is in the horizontal direction, therefore, considering the horizontal component of the velocity

Initial velocity in the x-direction,ux = ucosθ and final velocity(i.e at C) = vx =0 ,acceleration along x axis is 0

Applying second equation of the motion

s = ut + (1/2) at²

R = ucosθ.T + (1/2) ×0×T²

Substituting the value of T

R = ucosθ.2usinθ/g

R = u²2cosθsinθ/g

R = u²sin 2θ/g

Circular Motion: Angular velocity and angular displacement

Addition of Vectors: CBSE Class 11 Physics Chapter 4 -Motion in a Plane

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