krit.club logo

Gravitation - Free Fall and Acceleration due to Gravity

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Free Fall: When an object falls towards the Earth solely under the influence of gravitational force, with no other forces (like air resistance) acting on it, it is said to be in a state of free fall.

Acceleration due to Gravity (gg): During free fall, the object undergoes a uniform acceleration towards the center of the Earth. This acceleration is denoted by gg and its value on the surface of the Earth is approximately 9.8 m/s29.8 \text{ m/s}^2.

Calculation of gg: The value of gg is derived from the Universal Law of Gravitation. If MM is the mass of the Earth and RR is its radius, then g=GMR2g = \frac{GM}{R^2}.

Variation of gg: Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator. Since the radius RR is smaller at the poles, the value of gg is greater at the poles than at the equator.

Equations of Motion under Gravity: For objects moving under the influence of gravity, the standard equations of motion are used by replacing acceleration aa with gg and distance ss with height hh.

Sign Convention: When an object is falling downwards, gg is taken as positive (+9.8 m/s2+9.8 \text{ m/s}^2). When an object is thrown vertically upwards, gg is taken as negative (9.8 m/s2-9.8 \text{ m/s}^2) because the motion is against gravity.

Mass vs. Weight: Mass (mm) is the measure of inertia and remains constant everywhere. Weight (WW) is the force with which the Earth attracts an object (W=m×gW = m \times g) and varies depending on the value of gg at a location.

📐Formulae

g=GMR2g = \frac{G \cdot M}{R^2}

v=u+gtv = u + gt

h=ut+12gt2h = ut + \frac{1}{2}gt^2

v2=u2+2ghv^2 = u^2 + 2gh

W=mgW = m \cdot g

💡Examples

Problem 1:

A stone is dropped from the top of a building and reaches the ground in 3 s3 \text{ s}. Calculate the height of the building and the final velocity of the stone just before hitting the ground. (Take g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

Given: u=0 m/su = 0 \text{ m/s}, t=3 st = 3 \text{ s}, g=9.8 m/s2g = 9.8 \text{ m/s}^2. To find hh: h=ut+12gt2    h=0×3+12×9.8×(3)2    h=4.9×9=44.1 mh = ut + \frac{1}{2}gt^2 \implies h = 0 \times 3 + \frac{1}{2} \times 9.8 \times (3)^2 \implies h = 4.9 \times 9 = 44.1 \text{ m}. To find vv: v=u+gt    v=0+9.8×3=29.4 m/sv = u + gt \implies v = 0 + 9.8 \times 3 = 29.4 \text{ m/s}.

Explanation:

Since the stone is dropped, initial velocity uu is zero. We use the second equation of motion to find the distance (height) and the first equation to find the final velocity.

Problem 2:

An object is thrown vertically upwards and rises to a height of 20 m20 \text{ m}. Calculate the velocity with which the object was thrown upwards. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Given: h=20 mh = 20 \text{ m}, v=0 m/sv = 0 \text{ m/s} (at maximum height), g=10 m/s2g = -10 \text{ m/s}^2 (upward motion). Using v2=u2+2ghv^2 = u^2 + 2gh: 02=u2+2(10)(20)    0=u2400    u2=400    u=20 m/s0^2 = u^2 + 2(-10)(20) \implies 0 = u^2 - 400 \implies u^2 = 400 \implies u = 20 \text{ m/s}.

Explanation:

For upward motion, the final velocity at the highest point is zero and the acceleration due to gravity is taken as negative.

Free Fall and Acceleration due to Gravity Revision - Class 9 Science CBSE