# The projectile motion of an object thrown from a certain height : Class 11 Physics

**The projectile motion of an object thrown from a certain height:**If an object is thrown from a certain height by an angle θ with a velocity of u, the velocity u will have two components, horizontal u_{x} and vertical u_{y }, here u_{x} remains the same because the acceleration along the x-axis is zero while the vertical component u_{y }of velocity first decreases upto a certain height and then increases downwards achieves its maximum value nearest to the ground, the gravitational acceleration will always be negative because it has been applied on the vertical component of the velocity.

**The projectile motion of an object thrown from a certain height :**

**The projectile motion of an object thrown from a certain height:**Let an object is thrown from the top of a tower of the height h, the velocity of the object,u will have two component horizontal u_{x} and vertical u_{y }the path of the object is of parabolic shape because the object is dragged vertically and horizontally by both component of velocity, the vertical component will become 0 when it achieves a maximum height of H.

The displacement covered by the object in reaching from A to C is = -h

The time taken by the object in the journey is = T

Since displacement is along the y-axis.so let us take y component of velocity

Acceleration = -g(since it acts on the vertical component of velocity)

The initial velocity of the object = usinθ

Applying the second equation of the motion

s = ut +( 1/2)at²

-h =usinθ.T + ( 1/2)(-g)T²

( 1/2)(g)T²- usinθ.T -h =0

**gT²- 2usinθ.T -2h =0…….(i)**

Solving this quadratic equation, we can get the time taken by the object from the top of the building to the ground.

The range of the object is actually the displacement of the object in the direction of the x-axis

The time taken from the equation (i) is = T

The initial velocity of the object in the direction of the x-axis is = ucosθ

Acceleration along the x-axis is =0

The displacement along the x-axis = R

Applying the second equation of the motion

s = ut +( 1/2)at²

R = ucosθ.T + ( 1/2)×0×t²

**R = uTcosθ……(ii)**

Therefore we can calculate the range and time is taken by the object to reach the ground when it is thrown by an angle θ from a certain height from the ground

**The projectile motion of an object thrown from a certain height**

The final velocity of the object just before hitting the ground: The x component of the velocity remains the same even when it reaches to the ground at point C , the velocity which is changing is the vertical component, which was initially usinθ finally would become vsinθ

The acceleration along y-axis = -g

Applying the first equation of motion

v = u + at

vcosθ = ucosθ – gT

**v = (ucosθ -gT)/cosθ **

Finding the height of the tower when an object is thrown with a particular velocity of u downwards by an angle of θ from the horizon:

The height of the tower is =displacement in the direction of y axis =h

The initial velocity of the object is =-usinθ(- sign shows because the object is thrown in a downward direction)

The acceleration is = -g(since the displacement and gravitational acceleration are opposite to each other)

Displacement in the direction of x-axis is =R

Acceleration in the direction of x-axis is = 0

The initial velocity in the x direction is =ucosθ

Applying the second equation of the motion

s = ut +( 1/2)at²

R= ucosθ.T(since a =0)

T = R/ucosθ.

Applying the second equation of the motion

s = ut +( 1/2)at²

**h = usinθ.t -( 1/2)gT²**

**Projectile Motion:**A particle moves along a curved path under constant acceleration when thrown obliquely near the surface of the earth. This curved path is always directed towards the centre of the Earth. The path of such a particle is called the trajectory of the projectile, and the motion is called the projectile motion

We hope you would have liked the post ‘The projectile motion of an object thrown from a certain height ‘this topic is taken from the class 11 NCERT physics textbook which is very important to understand for solving the numerical of physics in class 11 CBSE Board exams.

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