## 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 *v*^{2} = u^{2 }– 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/s^{2})

**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

R_{1} = R_{2}

T_{1} ¹ T_{2}

H_{1} ¹ H_{2}

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 n^{th }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 50^{th} 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

u_{x }= u

u_{y} = 0

a_{x} = - g sin b

a_{y} = - g cos b

Now

Vx = u_{x} + a_{x}t

O = u – g sin bt

(ii) Final velocity V

V_{y }= u_{y} + a_{y} t

V = 0 – g cos b

V = u cot b

(iii) Height h

Now

h = S_{y} sin a

(iv) Distance AB

Now

**Case 2 **

(i) Time after which V_{1} ^ V_{2}

Ball 1

Ball 2

Now

g^{2 }t^{2} = u_{1 }u_{2}

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

Ball 1 *x*_{1 }= u_{1} t,

Ball 2

x_{2} = - u_{2}t

Gap between Ball 1 and Ball 2

Gap = (u_{1 }+ u_{2}) t

**Case 3 Collision of two projectile **