INTRODUCTION:One purpose of physics is to study the motion of objects how fast they move, for example, and how far they move in a given amount of time. NASCAR engineers are fanatical about this aspect of physics as they determine the performance of their cars before and during a race. Geologists use this physics to measure tectonic-plate motion as they attempt to predict earthquakes. Medical researchers need this physics to map the blood flow through a patient when diagnosing a partially closed artery, and motorists use it to determine how they might slow sufficiently when their radar detector sounds a warning. The world and everything in it, moves. Even seemingly stationary things, such as a roadway, move with Earth’s rotation, Earth’s orbit around the Sun and the Sun’s orbit around the center of the Milky Way galaxy, and that galaxy’s migration relative to other galaxies. The classification and comparison of motions (called kinematics) is often challenging. Medical researchers and aeronautical engineers might concentrate on the physics of the two- and three-dimensional turns taken by fighter pilots in dogfights because a modern high-performance jet can take a tight turn so quickly that the pilot immediately loses consciousness. A sports engineer might focus on the

physics of basketball. For example, in a free throw (where a player gets an uncontested shot at the basket from about 4.3 m), a player might employ the overhand push shot, in which the ball is pushed away from about shoulder height and then released. Or the player might use an underhand loop shot, in which the ball is brought upward from about the belt-line level and released. The first technique is the overwhelming choice among professional players, but the legendary Rick Barry set the record for free-throw shooting with the underhand technique.Motion in three dimensions is not easy to understand. For example, you are probably good at driving a car along a freeway (one-dimensional motion) but would probably have a difficult time in landing an airplane on a runway (three dimensional motions) without a lot of training.

DEFINITION:

A change from position x1 to position x2 is called a displacement ∆x. ∆x = x2 - x1.(The symbol ∆, the Greek uppercase delta, represents a change in a quantity, and it means the final value of that quantity minus the initial value.)Average velocity vavg is the ratio of the displacement ∆x that occurs during a particular time interval ∆t to that interval.

Average speed savg is a different way of describing “how fast” a particle moves. Whereas the average velocity involves the particle’s displacement ∆x, the average speed involves the total distance covered (for example, the number of meters moved), independent of direction; that is,

The velocity at any instant is obtained from the average velocity by shrinking the time interval ∆t closer and closer to 0.As ∆t dwindles, the average velocity approaches a limiting value, which is the velocity at that instant:

Where the particle has velocity v1 at time t1 and then velocity v2 at time t2. The instantaneous acceleration (or simply acceleration) is

If you tossed an object either up or down and could somehow eliminate the effects of air on its flight, you would find that the object accelerates downward at a certain constant rate. That rate is called the free fall acceleration, and its magnitude is represented by g. The acceleration is independent of the object’s characteristics, such as mass, density, or shape; it is the same for all objects.

The free-fall acceleration near Earth’s surface is a= -g= -9.8 m/s2, and the magnitude of the acceleration is g = 9.8 m/s2

PROJECTILE MOTION:

An important example of motion in a plane with constant acceleration is the projectile motion. When a particle is thrown obliquely near the earth’s surface, it moves along a curved path such a particle is called a projectile and its motion is a called projectile motion.

The total time taken by the particle in describing the path is called the time of flight.

Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s horizontal

acceleration is zero and its vertical acceleration is the free-fall acceleration -g. (Upward is taken to be a positive direction.) If is expressed as a magnitude (the speed v0) and an angle u0 (measured from the horizontal), the particle’s equations of motion along the horizontal x axis and vertical y axis are

The trajectory (path) of a particle in projectile motion is parabolic and is given by

If x0 and y0 is zero. The particle’s horizontal range R, which is the horizontal distance from the launch point to the point at which the particle returns to the launch height, is

The horizontal range R is maximum for a launch angle of 45°.

However, when the launch and landing heights differ, as in shot put, hammer throw, and basketball, a launch angle of 45° does not yield the maximum horizontal distance.

A special case of projectile:

APPLICATIONS OF KINEMATICS:

Case:

A helicopter on flood relief mission, flying horizontally with speed u at a height H, has to drop a food packet for a victim standing on the ground.

Application: At what distance from victim should the packet be dropped?

Height H can be equated to the maximum height of the projectile.

Then H= (u2

sin2@)/2g

@ is unkown

Then, tan@= 4H/R.

By solving it using mathematical formulas we can obtain the value of R/2 which is that distance D.

D=u (2H/g)1/2

Case:

A ball is hit by the batsman and so it gains

12 m/sec and it is at 45o initially. If the boundary length is 10m, will it be a six

Application:

Let us find the distance at which the ball will hit the field.

The horizontal range = (u2 sin2@)/g

Therefore the distance will be around 14.4 m

So the shot will be a sixer (ideally).

Case:

A war was going on, in which canons and copters were fighting for their countries.

Application:

To find the time period of danger.

Given that tan@=2 height of copter is y

Vcopter=v velocity of canon bombs=u

y=2x (1-x/R)

Ry=2xR – 2x2

X= (2R + (4R2-8yR)1/2)/2

Let a and b be the roots of this equation.

Then,

a - b= (R2 - 2yR)1/2

Time period = t = (a - b)/v

Therefore the time of danger is

t = (u/gv)*(u2 – 2gy)1/2

Figure shows a pirate ship 560 m from a fort defending a harbor entrance. A defense cannon, located at sea level, fires balls at initial speed v0 = 82 m/s. (a) At what angle u0 from the horizontal must a ball be fired to hit the ship?

(1) A fired cannonball is a projectile. We want an equation that relates the launch angle u0 to the ball’s horizontal displacement as it moves from cannon to ship. (2) Because the cannon and the ship are at the same height, the horizontal displacement is the range. As the pirate ship sails away, the two elevation angles at which the ship can be hit draw together, eventually merging at u0 =45° when the ship is 690 m away. Beyond that distance the ship is safe. However, the cannonballs could go farther if the cannon were higher.