PHYSICS IS EASY !: GOT NIGHT CLASS . . =="

assalamualaikum .
harinie kami ade kelas waktu malam . haiiihh . .bulan pon bagi semangat taw nak belajar mlm2 . haha .

tp x pe la . demi untuk belajar .kami pegi jugak .naseb la haritu mOOd baek .yg dtg lambat x perlu denda .
tq so much dr . =) ape yg kami belajar malam tue . ini diaaa . .tajuk baru chapter 3 .KINEMATICS .
dengar cerita masuk test kan . OMG . semoga kami dapat buat . .AMINNN . . nie la yg kami belajar . .

"Kinematics is the study of how things move. Here, we are interested in the motion of normal objects in our world. A normal object is visible, has edges, and has a location that can be expressed with (x, y, z) coordinates. We will not be discussing the motion of atomic particles or black holes or light.
We will create a vocabulary and a group of mathematical methods that will describe this ordinary motion. Understand that we will be developing a language for describing motion only. We won't be concerned with what is causing or changing the motion, or more correctly, the momentums of the objects. In other words, we are not concerned with the action of forces within this topic. "

ini ialah kinematic equation

dan ini

equation for linear motion .

$v = v_i + a \Delta t \,$ $s = s_i + v_i\Delta t + \tfrac{1}{2} a(\Delta t)^2 \,$

$s = s_i + \tfrac{1}{2} (v + v_i)\Delta t \,$

$v^2 = v_i^2 + 2a(s - s_i) \,$

Equations for Vertical Motion

Case I:

When a body is projected vertically downwards with initial velocity u then the equation describing motions are

v = u + gt ……(1)

h = ut + 1/2 gt² ……(2)

v² = u² + 2gh ……(3)

h_n = distance covered in n^th second = u + 1/2 g(2n –1)

Case II:

When body is falling down freely, then

v = gt ……(1)

h = 1/2 gt² ……(2)

v² = 2gh ……(3)

h_n = 1/2 g(2n –1) ……(4)

Case III:

When body is projected vertically upward with initial velocity u, then

v = u –gt ……(1)

h = ut –1/2gt² ……(2)

v² = u² –2gh ……(3)

h_n = u –1/2g(2n –1) ……(4)

Greatest height attained = u²/2g ……(5)

Time to the greatest height = u/g ……(6)

Time to a given height h = u/g ± √u²–2gh/g.

Notes:

Time from any point on the path to the highest point is same as the time from the highest point to the given point when the body is returning.

Solved Example 1:

A body is thrown vertically upward and rises to a height of 10 meters. Calculate (i) the velocity with which the body was thrown upwards and (ii) the time taken by the body to reach the highest point.

Solution:

Let the body be thrown upwards with a velocity of 4 m/sec. Thus greatest height attained = u²/g meters.

(i) We have, greatest height = 10 meters

=> u²/2g = 10 => u² = 196 => u = 14 m/sec

Hence, the required velocity of projection = 14 m/sec

(ii) Time taken by the body to reach the greatest height = u/g sec = 14/9.8 = 10/7 sec.

Solved Example 2:

The greatest height attained by a particle projected vertically upwards is 19.6 meters. Find how soon after projection the particle will be at a height of 14.7 meter.

Solution:

Let u be the velocity of projection since at the greatest height, the velocity of the particle is zero.

0² = u² –2g × 19.6 => u² = 19.6 × 19.6 => u = 19.6 m/sec

Suppose after time t the particle is at a height of 14.7 meter, then

14.7 = 19.6t – 1/2 × 9.8 × t²

= t² –4t+ 3 = 0 => (t –1) (t –3) = 0 => t = 1, 3

Thus, the particle will be at a height of 14.7 meter after 1 sec of its projection. It will also attain the same height after 3-seconds while coming downward after attaining the maximum height.

dan selepas itu kami di beri tugasan .

membuat sebanyak 22 soalan . for guys jwab soalan genap . n for girls jwab soalan ganjil .

soalan yg mencabar minda . not easy but take it easy . hahahaha . .

that's ALL . smoga rakan2 yang lain dpat menjawab soaln dgn jayanya . GOOD LUCK GUYS !

THANK YOU FOR READING .