## 1-D,2-D and 3-D motion, Trajectory(path) of projectile

1-D MOTION

2-D MOTION OR MOTION IN A PLANE Þ

Motion of any object in any of two axis involve (xy or yz or zx)

“Projectile Motion may be 2-D or even 3-D also but in general it is use to be 2-D Motion”

-D PROJECTILE MOTION

VERTICAL MIRROR

Vertical Mirror ®  Gravity acts in vertical direction so we can use equations of motion

Time taken by the image of ball going up = time taken going down = t (Let)

Use  v = u – g t

O = u sinq - g t

Total time

Now to get Maximum height

Using v2 = u2 – 2g h

(O)2 = (usinq)2 – 2g H

HORIZONTAL MIRROR

Velocity remains constant so we can not use equation of Motion

Note: Untill any external reason present to change ucosq, like air flow ucosq remains Constant

Now ucosq = Const

R = (u cosq) T

Now Sin2q = 2 sinq Cosq

Now we have

CONDITION

“These three results only when can be use if initial point of projection and final point of projection are at same level.

Example: A body of mass m is projected upward with initial velocity                                  then find time of flight, Maximum height attained and range attained by the body (g = 10m/s2)

Solution

Maximum range: To get maximum horizontal distance covered by any mass in projectile motion we must have a unique specified angle.

(Sin2q)max = 1

2q = 90°

q = 45°

To get maximum range angle of projection should be 45 °,

Note: Here we are neglecting the effect of air resistance.  If we Consider air resistance then this angle q should be little bit less than 45°

SAME RANGE

Mathematically there must be two different angle of projection for which we will get same range

Ball 1

Ball 2

Now if we assume

a + b = 90°

then b = 90° - a

If sum of angle of projection is = 90° and initial speed is same then both balls will have same range.

Note: Here in this Case only range will be same not time of flight and maximum height

R1 = R2

T1 ¹ T2

H1 ¹ H2

In case of same range relation of time of flights

Here we know a + b = 90°

Ball 1

Ball 2

Now

In case of same range relation of maximum heights

Here again a +b = 90°

Ball 1

Ball 2

=

## Path of Projectile

TRAJECTORY (PATH) OF PROJECTILE

Along y- axis  using

Along x-axis using

By this equation it can be prove that path of projectile is parabola

OTHER FORMA OF EQUATION OF TRAJECTORY

Projectile Motion when initial point and final point are at different level

Case I Horizontal projection from certain height

Here angle of projection q = 0˚

In vertical Mirror

Using

Time to strike ground in horizontal Mirror

IN HORIZONTAL MIRROR

=

Example: Which ball will strike the ground first

(a) Ball A          (c) Ball C

(b) Ball B         (d) All Balls will strike simultaneously

Solution

There fore all three Balls strike the ground simultaneously                      Answer (d)

Example: A jet fighter plane is flying at constant speed of 360 km/h and jet plane is moving horizontally at a height of 20 km from earth ‘s surface. Jet release a bomb to hit  a target on the ground, how much horizontal gap before jet will release the bomb

Solution

As discussed above using

Example: A stair case having each step 10 cm high and 20 cm wide.

If a ball drops in horizontal direction from the top step with 10 mt/sec

in horizontal direction If ball strikes nth step of staire case find the n?

Solution

Now make a twist in above question

Every  data is same except velocity of ball, if initial velocity not given , and saying what will be minimum initial velocity in above question so that ball hits 50th step

Solution

Now x = 49 x 0.2 mt

h = 49 x 0.1 mt

## Projection in upwards and downwards directions

Case II Projection in upward direction with some angle with horizonta

Vertical mirror

sign convention

Horizontal Mirror

x = (u cosq) t

Case III Projection in downward direction with some angle with horizontal

Vertical Mirror

Horizontal Mirror

Projectile along inclined plane For Simplicity & understanding

Vertical Mirror

V = u - g t

O = u sin b - (g cosa) t

Horizontal Mirror

OB = OA Cos a

Special case of Projectile Motion

Case I

(i) Time of flight  let OQ ® x -axis

OP ® y –axis

ux = u

uy = 0

ax = - g sin b

ay = - g cos b

Now

Vx = ux + axt

O = u – g sin bt

(ii) Final velocity V

Vy = uy + ay t

V = 0 – g cos b

V = u cot b

(iii) Height h

Now

h = Sy sin a

(iv) Distance AB

Now

Case 2

(i) Time after which V1  ^ V2

Ball 1

Ball 2

Now

g2 t2 = u1 u2

(ii) gap between the ball 1 and ball 2 at the time

Ball 1   x1 = u1 t,

Ball 2

x2 = - u2t

Gap between Ball 1 and Ball 2

Gap  = (u1 + u2) t

Case 3 Collision of two projectile