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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