Kaysons Publication
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
v2 = u2 + 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
v2 = u2 – 2gh
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 tth 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 5th sec and 6th 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,
5t2 –ut + 80 = 0
Now t must have two values t1 & t2
Using Sri Dharacharya
5t2 – ut + 80 = 0
Compare with ax2 + bx + c = 0
a = 5, b = - u, c = 80
Now t2 – t1 = 6
u2 -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
h1 : h2 : h3 …. 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: ….
= 12 : 22 : 32….. 42: 52
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/s2).
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/s2. He reaches the ground with a speed of 3 m/s At what height did he bail out. (g = 9.8 ms2)
Important Case (6):- Parachutist
Example: A parachutist after bailing out tells 50 meter without friction when parachute opens, it decelerates at 2m/s2. He reaches the ground with a speed of 3 m/s At what height did he bail out. (g = 9.8 ms2)
Solution:-
V2 = u2 + 2 gh
V2 = 2 x 9.8 x 50
Now after parachute opens
V2 = V2 -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
5t2 – 7t - 60 = 0
Compare ax2 + 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/s2 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
Vmax = 60 m/s
to get total distance moved by the train we will get total area under the curve.
A1 + A2 = 2700 mt
= 2.7 km Ans.
