### Chapter : Projectile Motion

__Projectile Thrown From The Ground Level __

__Projectile Thrown From The Ground Level__

**Introduction**

**(i)**Any particle, which once thrown, moves freely in gravitational field of the earth, is defined as a projectile.

**(ii)**It is an example of two dimensional motion with constant acceleration.

**(iii)**Parabolic motion can be considered as two simultaneous motions in mutually perpendicular directions viz.

**(a)**Horizontal and

**(b)**Vertical

**1. Projectile Thrown From The Ground Level**

**(i)**The particle is thrown from the ground level at an angle q from the horizontal velocity u.

**(ii)**Initial velocity can be resolved into two components

u cos θ = Horizontal component

u sin θ = Vertical component

From Horizontal Component we obtain | From Vertical Component we obtain |
---|---|

(a) Range (b) Velocity along horizontal direction after time t(c) Horizontal displacement after time t | (a) Maximum height (b) Time of flight (c) Time taken to reach maximum height (d) Vertical displacement after time t (e) Vertical velocity |

**Special Note :**

The horizontal component of velocity (u cos θ) remains constant where as the vertical component changes constantly due to acceleration due to gravity 'g'.

In Horizontal Direction | In Vertical Direction |
---|---|

(a) Initial Velocity u_{x}= u cos θ (b) Acceleration = 0 (c) Velocity after time t v_{x}= u cos θ (d) Horizontal displacement after time t, x = u_{x}.t = u cos θ t | (a) Initial Velocity u_{y} = u sin θ (upward) (b) Acceleration a = g (downward) (c) Velocity after time t, v_{y} = u_{y} – gt = u sin θ – gt (d) Vertical displacement after time t, y = u _{y}t – 1/2 gt^{2} = u sin θ t – (1/2) gt ^{2} (e) Velocity at height h above the surface v _{y}^{2} = u_{y}^{2} – 2gh= u ^{2} sin2 θ – 2ghvy = ± (one going upwards and other going downwards) |

**1.1 Velocity at a general point P(x, y) :**

The direction of v from horizontal

**1.2 Displacement :**

= x+ y ⇒

In fig. OP gives the displacement of projectile in the position (x,y) tan b = y/x, b is the angle made by displacement with horizontal in the position (x, y)

**1.3**

**Trajectory equation :**

y = u sin θ t – (1/2) gt

^{2}and x = (u cos θ)t

From these equations, (eliminating t)

**1.4**

**The maximum height reached by the projectile :**

(u

_{y}= vertical velocity)

**1.5 Time of flight of the projectile :**

Time taken to reach max. height

(using v = u + gt)

The time interval from initial launch of projectile upto its return to the ground level is known as the time of flight (T) of projectile

where u

_{y}= vertical velocity

**1.6 Horizontal Range of The Projectile :**

The horizontal distance covered by the projectile during its time of flight is known as the horizontal range of the projectile.

(using v

^{2}= u

^{2}+ 2gh)

**(i)**For maximum range, θ = 45º, R

_{max}= u

^{2}/g

In this situation, H

_{max}=

**(ii)**The range of the projectile is same when its angle of projection from the horizontal is either q or (90º – θ).

**(iii)**Also

where u

_{x}= u cos θ = horizontal velocity

u

_{y}= u sin θ = vertical velocity

**1.7 The velocity of the projectile at any instant of time t :**

and its direction from horizontal

**Note :**The trajectory of the missiles, targeted at very long ranges is not a parabola rather than it is an ellipse.

**1.8 Change in Momentum :**

**(i)**Initial velocity = u cos θ + u sin θ

**(ii)**Final velocity = u cos θ– u sin θ

Change in velocity for complete motion

Δ ==– = – 2u sin θ

**(iii)**Change in momentum for complete motion Δ = –

= m (–) = m (– 2 u sin θ) = –2 m u sin θ

**(iv)**Velocity at the heighest point == u cos θ change in momentum at highest point

(m– m) = m [u cos θ– (u cos θ + u sin θ)] = – m u sin θ

**(v)**Kinetic energy at highest point = E

_{0}cos

^{2}θ where E

_{0}= Initial K.E.

**(vi)**Magnitude of velocity at height 'h' , by energy conservation law

⇒ (1/2) mu

^{2}+ 0 = (1/2) mv

^{2}+ mgh

⇒

**Special Note :**

**(a)**The direction of acceleration of the projectile is different from its velocity direction.

**(b)**The acceleration of the projectile remains constant in magnitude and direction that is why the path of its motion is parabola.

__Trending Articles & Blogs__

- Physics Tutor, Math Tutor Improve Your Child’s Knowledge
- How to Get Maximum Marks in Examination Preparation Strategy by Dr. Mukesh Shrimali
- 5 Important Tips To Personal Development Apply In Your Daily Life
- Breaking the Barriers Between High School and Higher Education
- 14 Vocational courses after class 12th
- Tips to Get Maximum Marks in Physics Examination
- Get Full Marks in Biology Class 12 CBSE

Download Old Sample Papers For Class X & XII

Download Practical Solutions of Chemistry and Physics for Class 12 with Solutions