**Equation of Motion:**

Condition: If acceleration is Constant

** Motion Under Gravity**

If any mass is throwing in air then a downward gravitational force mg acts on the mass and this type of motion is known as motion under gravity.

**Case I:- **When ball is dropping from the top equations of motion.

V = u + gt

*v*^{2} = *u*^{2} + 2 *gh*

Ball drops from any height then its initial velocity.

*u* = 0

**Case II:- **When ball is throwing upward with initial velocity u.

Equations of motion ,

*v* = *u* – *gt*

*v*^{2} = *u*^{2} – 2*gh*

At the top when ball reaches its maximum height then it final velocity.

*V* = 0

**Important case (1) :- **

When a ball thrown vertical upward with velocity u and it is said that distance covered by ball in *t*^{th} sec is equal to (*t* + 1)^{th} sec.

Þ It implied that final velocity of ball *v* becomes zero after time *t*.

**Example:**

If a ball is throw with initial velocity u and it distance travelled by the ball in 5^{th} sec and 6^{th} sec is equal then what will be the maximum height covered by the ball.

**Solution: **

It is clear that time taken for upward motion of the ball is 5 sec there for down ward motion time will be 5 sec also.

For down ward motion

= 125 meter Ans

**Important Case (2):- **

If any object thrown vertically upward and it crosses certain height two time in a fixed interval of time.

**Example: **A ball is thrown upward from the ground with an initial speed u. The ball is at a height of 80 meter at two times, the time interval being 6 sec. Find value of u.

**Solution:-**

Now,

5t^{2} –ut + 80 = 0

Now t must have two values t_{1} & t_{2}

**Using Sri Dharacharya**

5t^{2} – ut + 80 = 0

Compare with *ax*^{2} + *bx* + *c* = 0

*a* = *5, b *= - *u*,* c *= 80

Now t_{2} – t_{1} = 6

u^{2} -1600 = 900

u = 50m/s

**Important Case (3):- **

If any object starting from rest and moving with uniform acceleration has distance covered in equal interval of time will be in ratio of 1:3:5:7……….

**Example:-**

If body drops from a certain height and it covers h distance in first 5sec then distance covered in next 5 sec also in further next 5 sec will be = ?

**Solution:-** fixed time interval is 5 sec by the ratio:-

For each 5 sec

h_{1} : h_{2} : h_{3} …. 1: 3: 5 : 7……

**Important Case:- (4) **

If a body starting from rest and moving with uniform acceleration then ratio of distance covered in 1sec, 2sec, 3sec….. = 1: 4 : 9: ….

= 1^{2} : 2^{2} : 3^{2}….. 4^{2}: 5^{2}

**Important Case: (5): **Juggler’s moving balls**.**

**Example: **A juggler** **keeps on moving four balls in the air throws the balls in regular interval of time. When one ball leaves his hand with speed 20m/s the position of other balls will be (g = 10m/s^{2}).

**Solution:-**

V = u – gt

O = 20 – 10t

t = 2 sec for each ball to reach max

Now for ball 1

= 5 mt

From top so 15 mt from bottom

**Maximum height**

= 40 - 20

= 20 mt

Ball 1 is at 5 mt

5mt, 15mt, 20mt

**Important Case (6):- Parachutist **

**Example**: A parachutist** **after bailing out fells 50 meter without friction when parachute opens, it decelerates at 2m/s^{2}. He reaches the ground with a speed of 3 m/s At what height did he bail out. (g = 9.8 ms^{2})

**Important Case (6):- Parachutist **

**Example**: A parachutist** **after bailing out tells 50 meter without friction when parachute opens, it decelerates at 2m/s^{2}. He reaches the ground with a speed of 3 m/s At what height did he bail out. (g = 9.8 ms^{2})

**Solution:- **

V^{2} = u^{2} + 2 gh

V^{2} = 2 x 9.8 x 50

Now after parachute opens

V^{2} = V^{2} -2a (H – h)

9 = 980 – 2 x 2 (H - h)

4 (H - h) = 980 – 9

H = 242.75 + h = 242.75 + 50

≈ 293 mt. **Ans**

**Important Case:- (7)**

Ball drops from rising balloon.

Example:- A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 met at the time of dropping the ball, how long will the ball take in reaching the ground

**Solution:- **** **

Using

Sigh convention

5t^{2} – 7t - 60 = 0

Compare ax^{2} + bx + c = 0

Taking + sigh

**Graph of kinematics**

**Examples:-**

(1) A body is moving with constant speed (v) and covers distance (s) in time (t)

(2) A body starts from rest and moves with constant acceleration (a) attained velocity (Ѵ) and covers a displacement (s) in time t.

Ѵ = u + at

Ѵ = o + at

Ѵ = at

By y = mx** **

(3) A body is moving with speed u then accelerate with constant acceleration (a) and attain speed *u* as well as covered distance (s) in time t

**Example based on graph**

A train starts from rest and moves with a constant acceleration of 2 m/s^{2} for half a minute. The breaks are then applied and trains comes to rest in one minute.

Find (i) Total distance moved by the train.

(ii) The maximum speed attained by the train

**Solution:** He we will draw V-t graph

Now slope of V-t group Þ acceleration

V_{max} = 60 m/s

to get total distance moved by the train we will get total area under the curve.

A_{1} + A_{2} = 2700 mt

= 2.7 km ** Ans.**