Mix Example - GRAVITATION Questions in English

(A) The gravitational force exerted by the earth on any object is specifically known as the weight of that object. It is defined as the force with which the earth attracts a body towards its center.

(N/A) Free fall is a state in which an object falls towards the Earth solely under the influence of gravitational force,with no other forces (like air resistance) acting upon it. During free fall,the object experiences a constant acceleration known as the acceleration due to gravity,denoted by $g$,which is approximately $9.8 \ m/s^2$ near the Earth's surface.

(N/A) One common phenomenon related to the universal law of gravitation is the motion of planets around the Sun. According to Newton's law of universal gravitation,every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This gravitational force provides the necessary centripetal force that keeps planets in their respective orbits around the Sun.

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses ($M_1$ and $M_2$) and inversely proportional to the square of the distance $(d^2)$ between their centers.
Mathematically,this is expressed as: $F = G \frac{M_1 M_2}{d^2}$,where $F$ is the gravitational force and $G$ is the universal gravitational constant.

(N/A) The force of gravity acts only in the vertically downward direction.
Since there is no force acting in the horizontal direction,there is no acceleration in the horizontal component of motion.
According to Newton's first law of motion,in the absence of an external horizontal force,the horizontal velocity remains constant throughout the flight.

(N/A) Yes, the statement is true. According to Newton's $Third$ $Law$ of $Motion$, the Earth exerts a gravitational force on the apple, and the apple exerts an equal and opposite force on the Earth. According to Newton's $Second$ $Law$ of $Motion$ $(F = ma)$, the acceleration produced is inversely proportional to the mass of the object $(a = F/m)$. Since the mass of the Earth is extremely large compared to the mass of the apple, the acceleration produced in the Earth is negligible and therefore not noticeable.

(A) The mass of an object is the quantity of matter contained in it.
Mass is a scalar quantity and remains constant regardless of the object's location in the universe.
Unlike weight,which depends on the acceleration due to gravity $(g)$,mass does not change with position.
Therefore,if the mass of the body is $10 \,kg$ on the surface of the earth,it will remain $10 \,kg$ at the centre of the earth.

(C) If the force of gravity disappears suddenly,all objects will lose the centripetal force required to maintain their circular motion around the Earth. According to Newton's first law of motion,objects will continue to move in a straight line tangent to their current path at the point where gravity vanished.

(N/A) The force of gravity acts in the vertically downward direction.
According to Newton's first law of motion,an object continues to move with a constant velocity unless acted upon by an external force in that direction.
Since the force of gravity acts only vertically downwards,there is no force acting in the horizontal direction to change the horizontal velocity.
Therefore,the horizontal component of the velocity remains constant throughout the motion.

(N/A) Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building. Observe whether both of them reach the ground simultaneously. We see that the paper reaches the ground a little later than the stone.
This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the paper is more than the resistance offered to the stone due to its larger surface area. If we perform the experiment in a vacuum (a glass jar from which air has been removed),the paper and the stone will fall at the same rate and reach the ground at the same time. This proves that acceleration due to gravity $(g)$ is independent of the mass of the object.

(N/A) $(i)$ According to Newton's law of universal gravitation,every object in the universe attracts every other object with a force that is proportional to the product of their masses.
$(ii)$ The earth exerts a gravitational force on the stone,causing it to accelerate towards the earth.
$(iii)$ The stone also exerts an equal and opposite gravitational force on the earth.
$(iv)$ However,due to the extremely large mass of the earth compared to the stone,the acceleration produced in the earth is negligible,making its motion towards the stone imperceptible.

(N/A) $(i)$ The weight on Mars is $38 \%$ of the weight on Earth.
Weight on Mars $= 0.38 \times 750 \, N = 285 \, N$.
$(ii)$ The relationship between weight $(W)$,mass $(m)$,and acceleration due to gravity $(g)$ is given by $W = m \times g$.
Given $W = 750 \, N$ and $g = 10 \, m s^{-2}$.
Mass $m = W / g = 750 / 10 = 75 \, kg$.
Therefore,Mona's mass on Earth is $75 \, kg$.

(N/A) $(i)$ The stone falls towards the earth because of the gravitational force exerted by the earth on the stone. According to the universal law of gravitation,every object in the universe attracts every other object with a force. Since the earth is massive,its gravitational pull on the stone is significant,causing the stone to accelerate towards it.
$(ii)$ According to the universal law of gravitation,the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Since the masses of ordinary objects in a room are very small,the gravitational force between them is extremely weak. This force is negligible compared to the frictional forces and the gravitational pull of the earth,which is why they do not move towards each other.

(N/A) Mass is defined as the quantity of matter contained in an object and is an intrinsic property that remains constant regardless of location.
Weight is defined as the force with which an object is attracted towards the center of a planet or celestial body,calculated as $W = m \times g$,where $m$ is mass and $g$ is the acceleration due to gravity.
Since the value of $g$ varies from place to place (e.g.,it is different on the Earth's surface compared to the Moon),the weight of an object changes accordingly,even though its mass remains constant.

(N/A) $1$. The gravitational force exerted by the Sun on the planets provides the necessary centripetal force for their motion along a circular orbit.
$2$. The direction of this force is always towards the center of the orbit (i.e.,towards the Sun).
$3$. If this force were to disappear suddenly,the planet would no longer be able to maintain its circular path and would continue to move in a straight line along the tangent to its orbit at that point,as per Newton's first law of motion.

(B) The formula for the force of gravitation is $F = G \frac{m_1 m_2}{r^2}$.
Given:
$m_1 = 10 \; kg$
$m_2 = 20 \; kg$
$r = 10 \; m$
$G = 6.67 \times 10^{-11} \; Nm^2 kg^{-2}$
Substituting the values into the formula:
$F = \frac{6.67 \times 10^{-11} \times 10 \times 20}{(10)^2} \; N$
$F = \frac{6.67 \times 10^{-11} \times 200}{100} \; N$
$F = 6.67 \times 10^{-11} \times 2 \; N$
$F = 13.34 \times 10^{-11} \; N$

(D) Given: Distance travelled by the stone,$S = 19.6 \, m$.
Initial velocity,$u = 0 \, m/s$.
Acceleration due to gravity,$a = 9.8 \, m/s^2$.
Using the third equation of motion: $v^2 - u^2 = 2aS$.
Substituting the values: $v^2 - 0^2 = 2 \times 9.8 \times 19.6$.
$v^2 = 19.6 \times 19.6$.
$v = \sqrt{19.6^2} = 19.6 \, m/s$.
Therefore,the velocity of the stone just before touching the ground is $19.6 \, m/s$.

(N/A) Mass of the object,$m = 50 \, kg$.
Acceleration due to gravity on the Earth,$g = 9.8 \, m/s^2$.
Gravitational force on the imaginary planet is $6$ times the gravitational force on the Earth.
Therefore,$g_{planet} = 6 \times g_{earth}$.
$g_{planet} = 6 \times 9.8 \, m/s^2 = 58.8 \, m/s^2$.
Weight of the object on the planet,$W = m \times g_{planet}$.
$W = 50 \, kg \times 58.8 \, m/s^2 = 2940 \, N$.
Thus,the acceleration due to gravity on the planet is $58.8 \, m/s^2$ and the weight of the object is $2940 \, N$.

(A) The weight of the man on the Earth's surface is $W_{E} = 600\, N$.
The mass of the man is calculated as $m = \frac{W_{E}}{g_{E}} = \frac{600\, N}{10\, m s^{-2}} = 60\, kg$.
Mass is a constant quantity and does not change with location,so the mass on the Moon remains $60\, kg$.
The weight of the man on the Moon's surface is $W_{M} = \frac{W_{E}}{6}$,because the acceleration due to gravity on the Moon is $\frac{1}{6}$ of that on the Earth.
Therefore,$W_{M} = \frac{600\, N}{6} = 100\, N$.

(C) According to Newton's law of universal gravitation,the force of attraction between two objects is given by $F = G \frac{M_1 M_2}{r^2}$.
This force is mutual,meaning the earth exerts a force on the lead,and the lead exerts an equal and opposite force on the earth.
According to Newton's third law of motion,for every action,there is an equal and opposite reaction.
Therefore,the gravitational force exerted by the earth on $1 \,kg$ of lead is exactly equal in magnitude to the gravitational force exerted by the $1 \,kg$ of lead on the earth.

(B) The weight of an object is defined as the force with which it is attracted towards the center of a planet,calculated as $W = m \times g$.
Since the mass $(m)$ of the man remains constant everywhere,his weight depends on the acceleration due to gravity $(g)$.
The acceleration due to gravity on the moon is approximately one-sixth $(1/6)$ of the acceleration due to gravity on the earth.
Therefore,the weight of the man will decrease on the surface of the moon compared to his weight on the earth.

(N/A) The acceleration due to gravity $(g)$ is greater at the poles than at the equator because the Earth is flattened at the poles and bulges at the equator,making the distance from the center of the Earth to the surface smaller at the poles.
According to the equation of motion $s = ut + (1/2)gt^2$,if a body is dropped from rest $(u = 0)$,then $s = (1/2)gt^2$,which implies $t = \sqrt{2s/g}$.
Since the distance $(s)$ is the same and the acceleration due to gravity $(g)$ is higher at the poles,the time $(t)$ taken to reach the ground is smaller at the poles compared to the equator.

(A) The effective acceleration due to gravity $(g')$ is given by the formula $g' = g - \omega^2 R \cos^2 \lambda$,where $\omega$ is the angular velocity of the earth,$R$ is the radius of the earth,and $\lambda$ is the latitude.
If the earth stops rotating,$\omega = 0$,which means the term $\omega^2 R \cos^2 \lambda$ becomes zero.
Consequently,the effective gravity $(g')$ increases at all points except at the poles (where $\lambda = 90^\circ$ and $\cos 90^\circ = 0$).
Since weight $(W = mg)$ is directly proportional to the effective acceleration due to gravity,the weight of a body will increase at all locations on the earth's surface except at the poles,where it remains unchanged.

(A) Yes,both objects will reach the earth at the same time.
This is because the vertical motion of both objects is independent of their horizontal motion.
Both objects start with an initial vertical velocity of $u_y = 0 \ m/s$.
Both objects are subjected to the same constant acceleration due to gravity $(g)$ in the downward direction.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $s$ is the height of the tower $(h)$,$u = 0$,and $a = g$,we get $h = \frac{1}{2}gt^2$.
Solving for time,$t = \sqrt{\frac{2h}{g}}$.
Since $h$ and $g$ are the same for both objects,the time taken $(t)$ to reach the ground is identical for both.

(N/A) According to Newton's law of universal gravitation,every object in the universe attracts every other object with a force proportional to the product of their masses and inversely proportional to the square of the distance between them $(F = G \frac{M_1 M_2}{r^2})$.
While all bodies do exert a gravitational force on each other,the magnitude of this force is extremely small for objects of ordinary size because their masses are negligible compared to the mass of the Earth.
In contrast,the Earth has an enormous mass. Therefore,the gravitational force exerted by the Earth on any object near its surface is significantly larger than the mutual gravitational forces between those objects.
Consequently,the Earth's gravitational pull dominates,causing all objects to be attracted towards the centre of the Earth.

(N/A) Weightlessness is a state in which an object experiences no net gravitational force or feels as though it has no weight. This phenomenon occurs when a body is in a state of free fall under the influence of gravity alone,where the supporting force (normal force) becomes zero.

(N/A) $(i)$ The centripetal force required for the earth to revolve around the sun is provided by the gravitational force of attraction between the sun and the earth.
$(ii)$ According to the law of universal gravitation,the magnitude of this force depends on the mass of the sun $(M)$,the mass of the earth $(m)$,and the square of the distance $(r^2)$ between their centers. The formula is $F = G \frac{Mm}{r^2}$,where $G$ is the universal gravitational constant.

(N/A) Let $m_{1}$ and $m_{2}$ be the masses of the two bodies and $M$ be the mass of the earth.
Let the distance of the two bodies from the centre of the earth be $r$.
According to Newton's law of universal gravitation,the force $F$ exerted by the earth on a body of mass $m$ at a distance $r$ is given by $F = \frac{GMm}{r^{2}}$,where $G$ is the universal gravitational constant.
For body $A$,the force is $F_{1} = \frac{GMm_{1}}{r^{2}}$.
For body $B$,the force is $F_{2} = \frac{GMm_{2}}{r^{2}}$.
Given that the forces are equal,$F_{1} = F_{2}$.
Therefore,$\frac{GMm_{1}}{r^{2}} = \frac{GMm_{2}}{r^{2}}$.
By cancelling the common terms $\frac{GM}{r^{2}}$ from both sides,we get $m_{1} = m_{2}$.
Thus,the masses of the two bodies are equal.

(N/A) The buoyant force depends on the following two factors:
$1$. The volume of the object submerged in the fluid.
$2$. The density of the fluid.
$(b)$ According to Archimedes' principle, the buoyant force acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. Therefore, $\text{Buoyant force} = \text{Weight of the fluid displaced}$.

(A) According to Newton's law of universal gravitation,$F = G \frac{m_1 m_2}{r^2}$.
$(i)$ If the distance $r$ is reduced to half,i.e.,$r' = r/2$,the new force $F'$ is:
$F' = G \frac{m_1 m_2}{(r/2)^2} = G \frac{m_1 m_2}{r^2 / 4} = 4 \left( G \frac{m_1 m_2}{r^2} \right) = 4F$.
So,the force becomes $4$ times the original force.
$(ii)$ If the mass of one object becomes four times,i.e.,$m_1' = 4m_1$,the new force $F''$ is:
$F'' = G \frac{(4m_1) m_2}{r^2} = 4 \left( G \frac{m_1 m_2}{r^2} \right) = 4F$.
So,the force becomes $4$ times the original force.

(N/A) The formula for the gravitational force is $F = \frac{G M m}{R^2}$,where $G$ is the universal gravitational constant,$M$ is the mass of the Earth,$m$ is the mass of the object,and $R$ is the radius of the Earth.
$(b)$ The direction of acceleration due to gravity is always towards the centre of the Earth.

(B) The acceleration due to gravity $(g)$ depends on the following factors:
$(i)$ Latitude of the place: The value of $g$ is maximum at the poles and minimum at the equator due to the Earth's shape and rotation.
$(ii)$ Height from the surface of the Earth: The value of $g$ decreases as we move away from the surface of the Earth (altitude).

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically,$F = G \frac{M \times m}{d^2}$,where $F$ is the gravitational force,$M$ and $m$ are the masses of the two objects,and $d$ is the distance between them.
The $SI$ unit of the universal gravitational constant $G$ is $N \ m^2 \ kg^{-2}$.

(N/A) In air,the coin will touch the ground first because the air resistance acting on the paper is significantly higher due to its larger surface area relative to its mass,which slows it down.
In a vacuum,both the coin and the paper will touch the ground at the same time.
This happens because,in a vacuum,there is no air resistance. According to the law of gravitation,all objects fall with the same acceleration due to gravity $(g)$,regardless of their mass or shape.

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically,$F = G \frac{m_1 m_2}{r^2}$,where $G$ is the universal gravitational constant.
If the distance $r$ is tripled,the new distance becomes $r' = 3r$.
The new force $F'$ will be $F' = G \frac{m_1 m_2}{(3r)^2} = G \frac{m_1 m_2}{9r^2} = \frac{1}{9} F$.
Therefore,the force becomes $\frac{1}{9}$ times the original force.

(N/A) $(i)$ It is the weakest force in nature.
$(ii)$ It is always attractive in nature and acts between any two objects having mass.

(B) Given that the mass of the planet $M_P = M/2$ and the radius of the planet $R_P = R/2$,where $M$ and $R$ are the mass and radius of the Earth respectively.
The formula for acceleration due to gravity on the surface of a planet is $g = GM/R^2$.
For the given planet,the acceleration due to gravity $g_P$ is given by:
$g_P = \frac{GM_P}{R_P^2}$
Substituting the values $M_P = M/2$ and $R_P = R/2$:
$g_P = \frac{G(M/2)}{(R/2)^2} = \frac{GM/2}{R^2/4} = \frac{GM}{2} \times \frac{4}{R^2} = 2 \times \frac{GM}{R^2}$
Since $g = GM/R^2$,we get:
$g_P = 2g$
Therefore,the acceleration due to gravity on the surface of this planet is $2g$.

(B) The weight of the object is the force required to lift it against gravity.
Given mass $m = 100 \,g = 0.1 \,kg$.
Gravitational acceleration on Earth $g \approx 10 \,m/s^2$.
The force $F$ required is equal to the weight $W = m \times g$.
$F = 0.1 \,kg \times 10 \,m/s^2 = 1 \,N$.
Therefore,a force of at least $1 \,N$ is required to lift the object.

(N/A) $(i)$ Acceleration due to gravity is the acceleration produced in a body due to the gravitational pull of the Earth.
$(ii)$ The Earth is not a perfect sphere; its radius is greater at the equator and smaller at the poles. Since $g = \frac{GM}{R^2}$,the value of $g$ varies with the radius $R$,making it higher at the poles and lower at the equator.
$(iii)$ The acceleration of a freely falling body is $g = \frac{GM}{R^2}$. This expression does not depend on the mass of the falling object $(m)$. Therefore,all objects fall with the same acceleration regardless of their mass,taking the same time to reach the ground from a fixed height.
$(iv)$ $G$ is called the universal gravitational constant because its value remains the same throughout the universe. Its $SI$ unit is $N \ m^2 \ kg^{-2}$.

(N/A) $(i)$ We know $g = \frac{GM}{R^2}$. Thus,on the surface of the earth and very close to the surface,'$R$' (i.e.,the radius of the earth) remains constant; hence,'$g$' will be constant. However,at distances far away from the surface of the earth,when '$R$' changes,the value of '$g$' will no longer remain constant.
$(ii)$ No,during a free fall,all objects fall with the same acceleration because '$g$' is independent of the mass of the falling body. The value of '$g$' is approximately $9.8 \, m/s^2$.

(N/A) Acceleration due to gravity is the uniform acceleration produced in a body falling freely under the influence of the earth's gravitational pull,directed towards the center of the earth.
$(b)$ The average value of $g$ on the surface of the earth is $9.8 \, m/s^2$.
$(c)$ The value of $g$ depends on: $(i)$ The shape of the earth (it is greater at the poles and smaller at the equator),$(ii)$ Height above the surface of the earth (it decreases with altitude),and $(iii)$ Depth below the surface of the earth (it decreases as we move towards the center).

(N/A) Acceleration due to gravity is the uniform acceleration produced in a body falling freely towards the surface of the earth due to the gravitational pull of the earth.
No,the acceleration produced in a freely falling body does not depend on the mass of the body.
According to Newton's second law of motion,the force $F$ acting on a body of mass $m$ is given by:
$F = m \times g$ --- $(1)$
According to Newton's law of universal gravitation,the gravitational force $F$ between the earth (mass $M$) and the body (mass $m$) at a distance $R$ (radius of earth) is:
$F = \frac{G M m}{R^2}$ --- $(2)$
Equating $(1)$ and $(2)$:
$m \times g = \frac{G M m}{R^2}$
Dividing both sides by $m$:
$g = \frac{G M}{R^2}$
Here,$G$ is the universal gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth. Since the expression for $g$ does not contain the mass of the falling body $(m)$,it proves that the acceleration due to gravity is independent of the mass of the body.

(A) The weight of an object is given by the formula $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.
The value of $g$ depends on the radius of the Earth $(R)$ as $g = \frac{GM}{R^2}$.
The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator. Therefore,the radius of the Earth at the equator is greater than the radius at the poles.
Since $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,the value of $g$ is lower at the equator and higher at the poles (like Antarctica).
Consequently,the weight of the bag of sugar will be more when taken to Antarctica compared to its weight at the equator.

(N/A) $(i)$ Weightlessness is observed in a satellite because the satellite and all objects inside it are in a state of free fall towards the Earth. Since there is no external contact force (normal force) acting on the body to oppose gravity,the apparent weight becomes zero.
$(ii)$ Given:
Initial velocity $(u) = 15 \, m s^{-1}$
Final velocity at maximum height $(v) = 0 \, m s^{-1}$
Acceleration due to gravity $(g) = -9.8 \, m s^{-2}$
Using the third equation of motion: $v^2 - u^2 = 2gh$
$0^2 - (15)^2 = 2 \times (-9.8) \times h$
$-225 = -19.6 \times h$
$h = \frac{225}{19.6} \approx 11.48 \, m$
Thus,the maximum height attained is approximately $11.48 \, m$.

(N/A) $(i)$ Newton's law of gravitation is applicable to all objects in the universe,regardless of their size,mass,or distance,hence it is called the universal law of gravitation.
$(ii)$ The universal law of gravitation is significant because it successfully explains the following phenomena:
$(a)$ The gravitational force that binds us to the Earth.
$(b)$ The motion of the Moon around the Earth.
$(c)$ The motion of planets around the Sun.
$(d)$ The occurrence of tides in the oceans due to the gravitational pull of the Moon and the Sun.

(N/A) $(i)$ The force of attraction between two objects is directly proportional to the product of their masses, i.e., $F \propto m_{1} m_{2}$.
The force of attraction between two objects is inversely proportional to the square of the distance between them, i.e., $F \propto \frac{1}{r^{2}}$.
$(ii)$ Given:
Mass of the Earth, $M = 6 \times 10^{24} \, kg$
Mass of the satellite, $m = 6.5 \times 10^{20} \, kg$
Distance between them, $r = 3.35 \times 10^{6} \, km = 3.35 \times 10^{9} \, m$
Gravitational constant, $G = 6.7 \times 10^{-11} \, Nm^{2} kg^{-2}$
Using the formula $F = G \frac{Mm}{r^{2}}$:
$F = \frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 6.5 \times 10^{20}}{(3.35 \times 10^{9})^{2}}$
$F = \frac{261.3 \times 10^{33}}{11.2225 \times 10^{18}}$
$F \approx 23.28 \times 10^{15} \, N$ or $2.33 \times 10^{16} \, N$.

(N/A) Let us consider an object of mass $m$ near the surface of a planet of mass $M$ and radius $R.$
According to the universal law of gravitation,the gravitational force $F$ acting on the object is given by:
$F = \frac{GMm}{R^2} \dots (1)$
According to Newton's second law of motion,the force $F$ acting on an object of mass $m$ is:
$F = mg \dots (2)$
(where $g$ is the acceleration due to gravity).
By equating equations $(1)$ and $(2)$,we get:
$mg = \frac{GMm}{R^2}$
Dividing both sides by $m$,we obtain:
$g = \frac{GM}{R^2}$
Since the mass of the object $m$ cancels out,the acceleration due to gravity $g$ depends only on the mass of the planet $M$ and its radius $R$. Thus,the acceleration experienced by an object during free fall is independent of its mass.

(B) The acceleration due to gravity $(g)$ on the surface of the earth is given by the formula:
$g = \frac{GM}{R^2}$
where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.
If the earth shrinks to one-third of its present size,the new radius $R'$ becomes $\frac{R}{3}$.
The mass $M$ remains constant.
The new acceleration due to gravity $g'$ is:
$g' = \frac{GM}{(R')^2} = \frac{GM}{(\frac{R}{3})^2} = \frac{GM}{\frac{R^2}{9}} = 9 \times \frac{GM}{R^2}$
Since $g = \frac{GM}{R^2} = 9.8 \, m s^{-2}$,we have:
$g' = 9 \times 9.8 \, m s^{-2} = 88.2 \, m s^{-2}$.

(N/A) Given:
Distance travelled by the ball,$S = 122.5 \, m$
Initial velocity,$u = 0$
Acceleration,$a = g = 9.8 \, m s^{-2}$
$(i)$ If $t$ is the time taken by the ball to fall to the ground,then using the equation of motion $S = ut + \frac{1}{2}at^2$:
$122.5 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$122.5 = 4.9 \times t^2$
$t^2 = \frac{122.5}{4.9} = 25$
$t = \sqrt{25} = 5 \, s$
$(ii)$ Distance travelled by the ball in the first three seconds $(t = 3 \, s)$:
$S = ut + \frac{1}{2}at^2$
$S = 0 \times 3 + \frac{1}{2} \times 9.8 \times (3)^2$
$S = 4.9 \times 9 = 44.1 \, m$
$(iii)$ Velocity of the ball at the end of three seconds $(t = 3 \, s)$:
$v = u + at$
$v = 0 + 9.8 \times 3 = 29.4 \, m s^{-1}$

(N/A)

Acceleration due to gravity $(g)$	Universal gravitational constant $(G)$
$(i)$ It is the acceleration acquired by a body due to the earth's gravitational pull on it.	$(i)$ It is numerically equal to the force of attraction between two masses of $1\, kg$ each separated by a distance of $1\, m$.
$(ii)$ The value of $g$ is different at different places on the surface of the earth and varies from one celestial body to another.	$(ii)$ The value of $G$ is a universal constant,i.e.,its value is the same $(6.67 \times 10^{-11}\, Nm^2 kg^{-2})$ everywhere in the universe.
$(iii)$ It is a vector quantity.	$(iii)$ It is a scalar quantity.