PHYSICS PROJECT SUBMITTED BY: Rakishma.M XI-E

PROJECTILE MOTION

PROJECTILE MOTION PProjectile motion: a combination of horizontal motion with constant horizontal velocity and vertical motion with a constant downward acceleration due to gravity.

PProjectile motion refers to the motion of an object that is thrown, or projected, into the air at an angle. TThe motion of a projectile is determined only by the object’s initial velocity and gravity. TThe vertical motion of a projected object is independent of its horizontal motion. TThe vertical motion of a projectile is nothing more than free fall. TThe one common variable between the horizontal and vertical motions is time.

EXAMPLES OF PROJECTILE MOTION

PATH OF A PROJECTILE AA projectile moves horizontally with constant velocity while being accelerated vertically. A right angle exists between the direction of the horizontal and vertical motion; the resultant motion in these two dimensions is a curved path. The path of a projectile is called its trajectory. The trajectory of a projectile in free fall is a parabola.

MOTION OF OBJECTS PROJECTED AT AN ANGLE TThe horizontal distance travelled by a projectile is determined by the horizontal velocity and the time the projectile remains in the air. The time the projectile remains in the air is dependent upon gravity. IImmediately after release of the projectile, the horizontal force is withdrawn & the force of gravity begins to accelerate the projectile vertically towards the Earth’s center of gravity.

Final speed = initial speed (conservation of energy) Impact angle = - launch angle (symmetry of parabola)

x y Motion is accelerated Acceleration is constant, and downward a = g = -9.81m/s 2 The horizontal (x) component of velocity is constant The horizontal and vertical motions are independent of each other, but they have a common time a = g = m/s 2

TThe horizontal component (X component) of the velocity v x remains constant over time because there is no acceleration along the horizontal direction. TThe vertical component of acceleration is the force of gravitation.

AAt the peak of the trajectory, The vertical component of the velocity v y is zero. However, there is a horizontal component of velocity, v x, at the peak of the trajectory. TThus at this point, the projectile travels in the resultant direction and follows a parabolic path.

THE RANGE OF A PROJECTILE TThe horizontal distance travelled by a projectile from its initial position to the position where it reaches the ground is called the horizontal range of the projectile. TT the horizontal range of the projectile is given by the equation: R= v 2 sin2  g

Derivation for the range of a projectile

THE TIME OF FLIGHT &THE MAX HEIGHT OF A PROJECTILE TThe total time taken by a projectile to complete its motion is called the total time of flight. It is given by the equation : T = 2vsin // g The maximum height reached by a projectile is given by the equation : H max = v 2 sin 2  g

Projectile Motion – Final Equations EQUATION : Total time T = 2VSin g Horizontal range R= v 2 sin2 g Max height H = v 2 sin 2  g (0,0) – initial position, v i = v i [Θ]– initial velocity, g = -9.81m/s 2

v0v0 x y Motion of Objects ProjectedHorizontally

x y

x y

x y

x y

x y Motion is accelerated Acceleration is constant, and downward a = g = -9.81m/s 2 The horizontal (x) component of velocity is constant The horizontal and vertical motions are independent of each other, but they have a common time g = -9.81m/s 2

x y 0 Frame of reference: h v0v0 Equations of motion: X Uniform m. Y Accel. m. ACCL.a x = 0a y = g = m/s 2 VELC.v x = v 0 v y = g t DSPL.x = v 0 ty=h+ ½ g t 2 g

For Objects Shot Horizontally vv x constant  y negative  y = -height

Trajectory x = v t y = h + ½ g t 2 Eliminate time, t t = x/v y = h + ½ g (x/v) 2 y = h + ½ (g/v 0 2 ) x 2 y = ½ (g/v 0 2 ) x 2 + h y x h Parabola, open down v 01 v 02 > v 01

Total Time, Δt y = h + ½ g t 2 final y = 0 y x h t i =0 t f =Δt 0 = h + ½ g (Δt) 2 Solve for Δt: Δt = √ 2h/(-g) Δt = √ 2h/(9.81ms -2 ) Total time of motion depends only on the initial height, h Δt = t f - t i

Horizontal Range, Δx final y = 0, time is the total time Δt y x h Δt = √ 2h/(-g) Δx = v 0 √ 2h/(-g) Horizontal range depends on the initial height, h, and the initial velocity, v 0 Δx x = v 0 t Δ x = v 0 Δ t

HORIZONTAL THROW - Summary TrajectoryHalf -parabola, open down Total time Δt = √ 2h/(-g) Horizontal Range Δx = v 0 √ 2h/(-g) Final Velocity v = √ v h(-g) h – initial height, v 0 – initial horizontal velocity, g = -9.81m/s 2

APPLICATIONS OF PROJECTILE MOTION TThere are many applications of projectile motion as everything that goes above the ground is attracted by gravity and hence projectile motion occurs. Some of them are: AAny soldier who has to bomb a particular place using an airplane or a tank must calculate the velocity and angle of throw for the bomb to hit the target

SSecondly it is widely used by people who extinguish fire. People who has to extinguish fire in a little longer distance from their stay show the tubes in an angle so that the water hits the fire, thus extinguishing it.

NNext it is most used by sportsmen especially the javelin throw, shot put, discus and hammer throw etc. It is also used by men of archery and shooting..

IIt is also used when, sometimes, food packets are thrown from helicopters in times of intense famine the distance from which the packets are thrown is important. If it is not calculated correctly, the food packets may fall in some other place.

IIt has high application in launching rockets and missiles as they follow the path of a projectile & in firing cannon balls.

PROJECTILE MOTION - SUMMARY PProjectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration TThe projectile moves along a parabola. TThe concept of projectiles has various applications. Some of them includes the launching of missiles, in extinguishing fire, in firing cannon ball etc.