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SS2: PHYSICS - 1ST TERM

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  1. Scalars & Vectors | Week 1
    5 Topics
    |
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    |
    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
    |
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    |
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    |
    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics
    |
    1 Quiz



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Scientists carried out various researches on the motion of objects falling under gravity and were able to come up with an acceptable concept that all bodies falling under gravity freely to the earth moved with the same acceleration irrespective of their masses. This gravitational acceleration is the acceleration of an object caused by the force of gravitation.

Neglecting friction such as air resistance, all bodies accelerate in a gravitational field at the same rate relative to the centre of mass.

This acceleration is known as acceleration due to gravity denoted by g.

When an object is falling to the earth, it obeys the law of gravity and it is positive g (+g) but when it is an object moving vertically upward, it is against gravity hence, it is denoted as negative g (-g), therefore, for bodies falling freely under gravitational motion, a is replaced by g, hence, the equation of motion for the uniformly accelerated bodies becomes;

v = \( \scriptsize u \pm gt \)

h = \( \scriptsize ut \pm \normalsize \frac{1}{2} \scriptsize gt^2 \)

v2 = \( \scriptsize u^2 \pm 2gh \)

s = h

At maximum height, h, the velocity = 0.

The initial velocity is zero when an object is released from a height, h.

When an object strikes the ground, h = 0,

If an object falls from a height, h, u = 0

h = \( \scriptsize ut + \normalsize \frac{1}{2} \scriptsize gt^2 \)

h = \( \scriptsize 0 + \normalsize \frac{1}{2} \scriptsize gt^2 \)

t2 =  \( \frac{2h}{g}\)

t = \(\sqrt{ \frac{2h}{g}}\)

Example 1:

An orange falls from a tree of height 40m, calculate the time of fall and velocity with which it strikes the ground.

Solution:

h = 40cm, g= 10ms-2, u = 0, v= ?

h = \(\frac{1}{2} \scriptsize gt^2 \)

t = \(\sqrt{ \frac{2h}{g}}\)

t = \(\sqrt{ \frac{2 \: \times \: 40}{10}} \scriptsize = \sqrt {8}\)

t = 2.83Secs.

Velocity = u + gt

v = 0 + 10 x 2.83

  = 28.3ms-1.

Example 2:

A Tennis ball is being projected from the top of a cliff 100m high with a velocity of 20m/s.
Calculate
(a) the height reached by the tennis ball above the ground
(b) How long will it take before the ball hits the ground?
(g = 10m/s2)

max height e1607008152382

(a) The height reached by the ball from the top of the tower = x

u = 20, h = x, g = 10m/s2, v = 0, hmax = 100m

Time to reach maximum height can be obtained from v = u + gt

0 = 20 + (−10)t

t = 2s

x = \( \scriptsize ut \: + \: \normalsize \frac{1}{2} \scriptsize gt^2 \)

x = \( \scriptsize 20(2) \: + \: 0.5(-10)(2)^2 \)

x = \( \scriptsize 40 \: – \: 20 \)

x = 20m

height reached by the tennis ball above the ground = hmax

hmax = 100m + 20m = 120m

(b) time it takes to fall from max height to the ground = t

t = \(\sqrt{ \frac{2h}{g}}\)

t = \(\sqrt{ \frac{2\: \times \: 120 }{10}}\)

t = \(\sqrt{ \frac{240 }{10}}\)

t = \(\sqrt{ \scriptsize 24}\)

t = 4.9s

Total time = 4.9s + 2s = 6.9s

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