Topics
System of Coplanar Forces
- Resultant of concurrent forces
- Resultant of parallel forces
- non-concurrent Non-parallel system of forces
- Moment of force about a point
- Moment of force about Couples
- Varignon’s Theorem
- Force couple system
- Distributed Forces in plane
Center of Gravity and Centroid for Plane Laminas
- Centroid for Plane Laminas
Equilibrium of System of Coplanar Forces
- Condition of equilibrium for concurrent forces
- Condition of equilibrium for parallel forces
- Condition of Equilibrium for non-concurrent nonparallel general forces
- Condition of equilibrium for Couples
Types of Support
- Load Support
- Beams Support
- Determination of reactions at supports for various types of loads on beams
Analysis of Plane Trusses
- Analysis of Plane Trusses by Using Method of Joints
- Analysis of plane trusses By using Method of sections
Forces in Space
Resultant of Noncoplanar force systems
- Resultant of Concurrent Force System
- Parallel Force System
- Non-concurrent Non-parallel Force System
Equilibrium of Noncoplanar force systems
- Equilibrium of Concurrent force system
- Equilibrium of parallel force system
- Equilibrium of non-concurrent non-parallel force system
Friction
- Introduction to Laws of Friction
- Cone of Friction
- Equilibrium of Bodies on Inclined Plane
- Application to Problems Involving Wedges
- Ladders
Principle of Virtual Work
- Applications on Equilibrium Mechanisms
- Applications on pin jointed frames.
Kinematics of Particle
- Rectilinear Motion
- Velocity and Acceleration in Terms of Rectangular Co-ordinate System
- Motion Along Plane Curved Path
- Tangentialand Normal Component of Acceleration
- Motion Curves (a-t, v-t, s-t curves)
- Projectile Motion
Kinematics of Rigid Bodies
- Introduction to General Plane Motion
- Instantaneous Center of Rotation for the Velocity
- Velocity Diagrams for Bodies in Plane Motion.
Kinetics of a Particle
Force and Acceleration
- Introduction to Basic Concepts
- D’Alemberts Principle
- Equations of Dynamic Equilibrium
- Newton’s Second Law of Motion
Work and Energy
- Principle of Work and Energy
- Principle of Conservation of Energy
Impulse and Momentum
- Principle of Linear Impulse and Momentum
- Law of Conservation of Momentum
- Impact and Collision
- Introduction
- Definition: Projectile Motion
- Key Concepts
- Characteristics
- Trajectory Equation
- Significance
- Example
- Real-Life Examples
Introduction
Projectile motion is the movement of an object in flight after being thrown with some initial velocity. Any object in flight after being thrown with some velocity is called a projectile. We see projectile motion every day—when children throw stones at trees to get fruit, when a bowler delivers a ball in cricket, or when a basketball player throws a ball toward the basket. In projectile motion, objects move under the influence of Earth's gravitational field. The motion involves two velocity components: one horizontal and one vertical.
Definition: Projectile Motion
Any object in flight after being thrown with some velocity is called a projectile, and its motion is called projectile motion.
Key concepts
Velocity Components: The projectile has two components of velocity:
- Horizontal component (along x-axis)
- Vertical component (along y-axis)
Acceleration in Projectile Motion:
- Horizontal acceleration: ax = 0 (no force acts horizontally)
- Vertical acceleration: ay = -g (gravitational acceleration, downward)
Characteristics
- The horizontal component of velocity remains constant throughout the motion.
- The vertical component of velocity changes continuously due to gravity.
- The trajectory is a parabolic path.
- At maximum height, vertical velocity becomes zero (vy = 0).
- The time taken to go up equals the time taken to come down (motion is symmetrical).
- The horizontal and vertical components of velocity are independent of each other.
- Air resistance is neglected in the study of projectile motion.
- The motion is governed solely by gravitational acceleration.
Trajectory Equation
Step 1: Understanding Initial Velocity Components
When a projectile is thrown at an angle θ with the horizontal, the initial velocity u is split into two components:
- Horizontal: ux = u cosθ
- Vertical: uy = u sinθ
Step 2: Horizontal Motion
Since no force acts in the horizontal direction, the horizontal component of velocity (vx) remains constant throughout the motion. This means:
- vx = u cosθ = constant
- Horizontal displacement: sx = u cosθ · t
Step 3: Vertical Motion
The vertical motion is similar to motion under gravity. The vertical velocity decreases as the projectile rises and increases as it falls:
- At any time t: vy = u sinθ - gt
- Vertical displacement: sy = u sinθ · t - ½gt²
Step 4: Motion at Maximum Height
At the maximum height (point P), the vertical velocity becomes zero:
- vy = 0
- Time to reach maximum height: t₀ = (u sinθ)/g
- Maximum height: H = (u sinθ)²/(2g)
Step 5: Complete Flight
Due to symmetry, the total time in the air is:
-
T = 2t₀ = (2u sinθ)/g
The horizontal distance covered (range) is:
-
R = u² sin(2θ)/g
Significance
- Understanding real-world phenomena: Helps explain everyday occurrences like throwing, jumping, and shooting
- Predicting trajectories: Allows us to calculate where and when a projectile will land
- Engineering applications: Used in designing sports equipment, weapons, and flight paths
- Maximum range calculation: Shows that 45° gives the maximum horizontal range
- Independent components: Demonstrates that horizontal and vertical motions are independent
- Parabolic path: Explains why projectiles follow a curved, parabolic trajectory
- Practical applications: Essential for sports, military strategies, and space missions
Example
Problem:
A stone is thrown with initial velocity components of 20 m/s along the vertical and 15 m/s along the horizontal direction. Determine the position and velocity of the stone after 3 s. Find the maximum height and total horizontal distance travelled.
Given:
- Horizontal component: u cosθ = ux = 15 m/s
- Vertical component: u sinθ = uy = 20 m/s
- Time: t = 3 s
- g = 10 m/s²
Solution Steps:
1. Find velocity components at t = 3 s
- vx = u cosθ = 15 m/s (remains constant)
- vy = u sinθ - gt = 20 - 10(3) = 20 - 30 = -10 m/s
The negative sign means the vertical velocity is 10 m/s downward.
2. Find total velocity magnitude
-
v = √(vx² + vy²) = √(15² + 10²) = √(225 + 100) = √325 = 18.03 m/s
3. Find the direction of velocity
- tan α = vy/vx = 10/15 = 2/3
- α = tan⁻¹(2/3) = 33° 41' (below horizontal)
4. Find displacement components
- sx = u cosθ · t = 15 × 3 = 45 m
- sy = u sinθ · t - ½gt² = 20(3) - ½(10)(3)² = 60 - 45 = 15 m
Position after 3 s: The stone is 45 m horizontally and 15 m vertically from the initial position.
5. Calculate maximum height
-
H = (u sinθ)²/(2g) = (20)²/(2 × 10) = 400/20 = 20 m
6. Calculate maximum horizontal distance (range)
-
R = (2 · ux · uy)/g = (2 × 15 × 20)/10 = 600/10 = 60 m
Results:
- Velocity at 3 s: 18.03 m/s at an angle of 33° 41' below horizontal
- Position at 3 s: 45 m horizontal, 15 m vertical
- Maximum height: 20 m
- Horizontal range: 60 m
Real-Life Examples
- Sports: A footballer kicking a goal, a basketball player shooting toward the basket, or a cricket bowler delivering a ball all involve projectile motion
- Military applications: Bullets and artillery shells follow projectile paths
- Space missions: Satellites and rockets follow projectile motion principles
- Water fountains: Water jets from fountains follow parabolic paths
- Diving and jumping: Athletes performing dives or long jumps follow projectile trajectories
- Fireworks: Fireworks launched into the air follow projectile motion before exploding
- Javelin and shot put: These athletic events involve throwing projectiles at specific angles
