Free Fall and Acceleration due to Gravity Questions in English

(N/A) The gravity of the Earth attracts every object towards its centre. When an object is released from a height,it falls towards the surface of the Earth solely under the influence of gravitational force. This motion of the object is known as free fall.

(N/A) When an object falls towards the ground from a height,its velocity changes during the fall. This change in velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity,denoted by $g$. Its standard value on the surface of the Earth is approximately $9.8 \, m/s^2$.

(B) According to Newton's second law of motion,$F = ma$. The gravitational force acting on an object is $F = G \frac{Mm}{r^2}$. Equating these,$ma = G \frac{Mm}{r^2}$,which simplifies to $a = G \frac{M}{r^2}$. Here,$a$ is the acceleration due to gravity $(g)$,$G$ is the gravitational constant,$M$ is the mass of the Earth,and $r$ is the distance from the center of the Earth. Since $g$ depends only on the mass of the Earth and the distance from its center,it is independent of the mass of the falling object $(m)$. Therefore,in the absence of air resistance,all objects fall with the same acceleration,and heavy objects do not fall faster than light objects.

(B) According to the universal law of gravitation,the gravitational force $F$ exerted on an object of mass $m$ is given by the formula:
$F = \frac{G M m}{r^2}$
Where:
Mass of Earth,$M = 6 \times 10^{24} \,kg$
Mass of object,$m = 1 \,kg$
Universal gravitational constant,$G = 6.67 \times 10^{-11} \,Nm^2 \,kg^{-2}$
Radius of the Earth,$r = R = 6.4 \times 10^6 \,m$
Substituting the values into the formula:
$F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2}$
$F = \frac{40.02 \times 10^{13}}{40.96 \times 10^{12}}$
$F \approx 9.8 \,N$
Thus,the magnitude of the gravitational force is approximately $9.8 \,N$.

(A) When objects fall towards the Earth under the effect of gravitational force alone,they are said to be in free fall.
The acceleration produced in an object due to the Earth's gravitational force is known as the acceleration due to gravity or acceleration of free fall.
Its value is approximately $9.8\, ms^{-2}$,which is constant for all objects regardless of their mass.

(B) When a sheet of paper is crumpled into a ball,its surface area exposed to the air decreases.
Since air resistance depends on the surface area of the object moving through it,the crumpled ball experiences significantly less air resistance compared to the flat sheet of paper.
As a result,the crumpled ball falls faster than the flat sheet of paper.

(D) Given:
Initial velocity $u = 49\, m/s$
Final velocity at maximum height $v = 0\, m/s$
Acceleration due to gravity $g = -9.8\, m/s^2$ (acting downwards)
$(i)$ To find the maximum height $(h)$:
Using the third equation of motion: $v^2 - u^2 = 2gh$
$0^2 - (49)^2 = 2 \times (-9.8) \times h$
$-2401 = -19.6 \times h$
$h = 2401 / 19.6 = 122.5\, m$
$(ii)$ To find the total time of flight $(T)$:
Using the first equation of motion for ascent: $v = u + gt$
$0 = 49 + (-9.8) \times t$
$9.8t = 49$
$t = 5\, s$
Since the time of ascent equals the time of descent,the total time $T = 2 \times t = 2 \times 5 = 10\, s$.

(A) According to the equation of motion under gravity $v^2 - u^2 = 2gs$.
Where,
$u = \text{Initial velocity of the stone} = 0\, ms^{-1}$.
$v = \text{Final velocity of the stone}$.
$s = \text{Height of the tower} = 19.6\, m$.
$g = \text{Acceleration due to gravity} = 9.8\, ms^{-2}$.
Substituting the values in the equation:
$v^2 - 0^2 = 2 \times 9.8 \times 19.6$.
$v^2 = 19.6 \times 19.6 = (19.6)^2$.
$v = 19.6\, ms^{-1}$.
Hence,the velocity of the stone just before touching the ground is $19.6\, ms^{-1}$.

(50 M) Let the initial heights be $H_1 = 150 \, m$ and $H_2 = 100 \, m$.
Both objects are dropped from rest,so their initial velocity $u = 0 \, m/s$.
Both objects fall with the same acceleration due to gravity,$g$.
Distance covered by an object in time $t$ is given by $s = ut + \frac{1}{2}gt^2$.
Since $u = 0$,the distance covered by both objects in $t = 2 \, s$ is $s = \frac{1}{2}g(2)^2 = 2g$.
After $2 \, s$,the height of the first object from the ground is $h_1 = H_1 - s = 150 - 2g$.
The height of the second object from the ground is $h_2 = H_2 - s = 100 - 2g$.
The difference in their heights after $2 \, s$ is $h_1 - h_2 = (150 - 2g) - (100 - 2g) = 150 - 100 = 50 \, m$.
Since the acceleration $g$ is the same for both,the distance covered by both objects at any time $t$ is the same $(s = \frac{1}{2}gt^2)$.
Therefore,the difference in heights at any time $t$ is $(H_1 - s) - (H_2 - s) = H_1 - H_2 = 150 - 100 = 50 \, m$.
Thus,the difference in heights remains constant and does not vary with time.

(A) When an object is in free fall,its acceleration is equal to the acceleration due to gravity $(g)$,which is independent of the mass of the object.
Since both objects are falling freely near the surface of the moon,they will experience the same acceleration due to gravity $(g_{moon})$.
Because they start from the same initial conditions (assuming they are dropped from the same height at the same time),their velocities at any given instant will be identical.
Therefore,they will have the same velocities at any instant.

(B) The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator.
As a result,the distance from the center of the Earth to the surface is greater at the equator than at the poles.
The formula for acceleration due to gravity is $g = \frac{GM}{R^2}$,where $g$ is inversely proportional to the square of the radius $(R)$.
Since the radius $R$ is largest at the equator,the value of $g$ is least at the equator.

(D) The acceleration due to gravity $(g)$ is zero at the centre of the earth.
We know that weight $(W)$ is the product of mass $(m)$ and acceleration due to gravity $(g)$,given by the formula $W = m \times g$.
Since the acceleration due to gravity at the centre of the earth is $0$,the weight of the object at the centre of the earth will be $m \times 0 = 0$.

(NONE) Both stones will reach the ground at the same time. This is because the vertical motion of both stones is governed by the same gravitational acceleration $(g)$ and they are released from the same initial height $(h)$. The horizontal velocity imparted to the first stone does not affect its vertical motion or the time taken to fall under gravity,as described by the equation of motion $h = \frac{1}{2}gt^2$,where $t = \sqrt{\frac{2h}{g}}$.

(B) The acceleration due to gravity,denoted by $g$,is not uniform across the Earth's surface.
Due to the Earth's rotation and its equatorial bulge,the radius of the Earth is greater at the equator than at the poles.
Since $g = GM/R^2$,where $R$ is the radius of the Earth,the value of $g$ is smaller at the equator and larger at the poles.
Using the equation of motion $h = (1/2)gt^2$,we get $t = \sqrt{2h/g}$.
Since $g$ is smaller at the equator,the time $t$ taken to reach the surface will be greater at the equator compared to the poles.
Therefore,the packets will not take the same time; the packet dropped at the equator will take a longer time to reach the surface.

(A) The acceleration due to gravity $g$ on the surface of the earth is given by the formula: $g = \frac{GM}{R^2}$,where $M$ is the mass of the earth and $R$ is its radius.
From this,we can express the mass $M$ as: $M = \frac{g R^2}{G}$.
The density $D$ of the earth is defined as the mass per unit volume: $D = \frac{M}{V}$.
Assuming the earth is a sphere,its volume $V$ is given by: $V = \frac{4}{3} \pi R^3$.
Substituting the expressions for $M$ and $V$ into the density formula:
$D = \frac{g R^2 / G}{4/3 \pi R^3} = \frac{g R^2}{G} \times \frac{3}{4 \pi R^3}$.
Simplifying the expression,we get: $D = \frac{3g}{4 \pi G R}$.

(N/A) For an object dropped from height $h$,the time $t$ taken to reach the ground is given by the equation of motion $h = \frac{1}{2} g t^2$,where $g$ is the acceleration due to gravity.
For the first object: $h_1 = \frac{1}{2} g t_1^2 \implies t_1 = \sqrt{\frac{2h_1}{g}}$.
For the second object: $h_2 = \frac{1}{2} g t_2^2 \implies t_2 = \sqrt{\frac{2h_2}{g}}$.
The ratio of time taken is $\frac{t_1}{t_2} = \frac{\sqrt{2h_1/g}}{\sqrt{2h_2/g}} = \sqrt{\frac{h_1}{h_2}}$.
$(i)$ If one object is hollow and the other is solid,the ratio remains the same because the acceleration due to gravity $g$ is independent of the mass and density of the object.
(ii) If both are hollow,the ratio still remains the same for the same reason. In free-fall,the acceleration of an object is independent of its mass,shape,or size.

(N/A) Free fall is defined as the motion of an object falling solely under the influence of gravitational force,with no other forces such as air resistance acting upon it. In this state,the object experiences an acceleration equal to the acceleration due to gravity,denoted by $g$.

(B) An object falling freely towards the Earth's surface under the influence of gravitational force is a classic example. For instance,when a ball is dropped from a height,it accelerates towards the Earth due to the gravitational pull exerted by the Earth on the object.

(N/A) The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator. The acceleration due to gravity 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. Since the radius $R$ is smaller at the poles compared to the equator,the value of $g$ is inversely proportional to the square of the radius,making $g$ greater at the poles.

(B) The acceleration due to gravity 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 distance from the center of the Earth.
Since the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius $R$ is smaller at the poles compared to the equator.
Because $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,a smaller radius at the poles results in a higher value of $g$.
Conversely,the larger radius at the equator results in a lower value of $g$.

(C) During free fall,an object is influenced only by the gravitational force of the Earth. According to Newton's second law of motion,the acceleration produced in a body due to gravity is independent of the mass of the object. This acceleration is known as the acceleration due to gravity $(g)$,which is approximately $9.8 \ m/s^2$ near the surface of the Earth. Therefore,for all bodies,whether large or small,undergoing free fall,the acceleration remains constant.

(N/A) The acceleration due to gravity $(g)$ at the surface of the earth is given by the formula:
$g = \frac{GM}{R^2}$
Where:
$G$ is the universal gravitational constant.
$M$ is the mass of the earth.
$R$ is the radius of the earth.

(C) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{4}{3} \pi G \rho R$,where $\rho$ is the density and $R$ is the radius of the planet.
Since $G$,$\pi$,and $R$ are constant,we have $g \propto \rho$.
If the density $\rho$ is doubled,the new acceleration due to gravity $g'$ becomes $g' = \frac{4}{3} \pi G (2\rho) R = 2g$.
Therefore,the value of $g$ on the planet is doubled.

(B) The acceleration due to gravity $(g)$ experienced by a body falling towards the Earth is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the universal gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
Since this formula does not include the mass of the falling body $(m)$,the acceleration is independent of the mass of the body.
Therefore,all objects fall with the same acceleration regardless of their mass in the absence of air resistance.

(D) When objects are dropped from a height,they undergo free fall under the influence of gravity.
According to the laws of motion,the acceleration of an object in free fall is equal to the acceleration due to gravity $(g)$,which is approximately $9.8 \, m/s^2$.
This acceleration due to gravity is independent of the mass,size,or shape of the falling object.
Since both spheres are dropped from the same height at the same time,at any given point (such as $10 \, m$ above the ground),they will have the same velocity and consequently the same acceleration.
Kinetic energy $(1/2 mv^2)$,momentum $(mv)$,and potential energy $(mgh)$ all depend on the mass $(m)$ of the object,which is different for the two spheres. Therefore,these quantities will not be the same.

(D) The value of acceleration due to gravity $(g)$ depends on the following factors:
$1$. Shape of the Earth: The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator. Due to this,the radius at the poles is less than at the equator,making $g$ greater at the poles.
$2$. Rotation of the Earth: The rotation of the Earth creates a centrifugal force that reduces the effective gravity,especially at the equator.
$3$. Latitude: Since the Earth's radius varies with latitude,the value of $g$ changes accordingly.
$4$. Altitude or Height: As we move away from the surface of the Earth (increasing altitude),the value of $g$ decreases.
$5$. Depth: As we move inside the Earth (increasing depth),the value of $g$ decreases.
Therefore,the value of $g$ depends on the shape,rotation,latitude,altitude,and depth.

(B) The acceleration due to gravity 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 distance from the center of the Earth.
Since the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius $R$ at the poles is less than the radius $R$ at the equator.
Because $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,a smaller radius at the poles results in a higher value of $g$ compared to the equator.

(B) The statement is false.
The value of acceleration due to gravity $(g)$ is not constant on the surface of the earth.
Because the earth is not a perfect sphere (it is flattened at the poles and bulges at the equator),the distance from the center of the earth to the surface is less at the poles and more at the equator.
Since $g = GM/R^2$,the value of $g$ is inversely proportional to the square of the radius $(R)$.
Therefore,$g$ is maximum at the poles and minimum at the equator.

(A) The acceleration due to gravity $(g)$ at a distance $(r)$ from the centre of the Earth is given by the formula $g' = g(1 - d/R)$,where $d$ is the depth from the surface.
At the centre of the Earth,the depth $(d)$ is equal to the radius of the Earth $(R)$.
Substituting $d = R$ into the formula,we get $g' = g(1 - R/R) = g(1 - 1) = 0$.
Therefore,the acceleration due to gravity at the centre of the Earth is $0 \ m/s^2$.

(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) 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.

(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 $2940 \, N$.

(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) 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) 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) 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) 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.

(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 $2g$.

(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.

(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 $(t = 3 \, s)$:
$v = u + at$
$v = 0 + 9.8 \times 3 = 29.4 \, m s^{-1}$

(A) Free fall is defined as the motion of an object falling towards the Earth solely under the influence of gravitational force,with no other forces like air resistance acting upon it.
In a vacuum,there is no air resistance. The acceleration experienced by any object falling freely is the acceleration due to gravity $(g)$,which is approximately $9.8\,m/s^2$ near the Earth's surface.
Since the acceleration due to gravity $(g)$ is independent of the mass of the falling object,both bodies will experience the same acceleration.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ (initial velocity) and $s = h$ (height),we get $h = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2h}{g}}$.
Since $h$ and $g$ are the same for both bodies,the time taken $(t)$ will be identical for both. Therefore,both bodies will hit the ground at the same time.

(N/A) For upward motion:
$v = u - gt$
At the highest point,final velocity $v = 0$.
So,$0 = u - gt_1$,which gives $t_1 = \frac{u}{g}$ $....(1)$
For downward motion:
The body starts from rest at the highest point,so initial velocity $u' = 0$.
Using $s = ut + \frac{1}{2}at^2$,where $s = h$ (maximum height) and $a = g$:
$h = 0 + \frac{1}{2}gt_2^2 \implies t_2 = \sqrt{\frac{2h}{g}}$.
Since $h = \frac{u^2}{2g}$,substituting this gives $t_2 = \sqrt{\frac{2(u^2/2g)}{g}} = \sqrt{\frac{u^2}{g^2}} = \frac{u}{g}$ $....(2)$
Comparing $(1)$ and $(2)$,$t_1 = t_2$. Thus,the time of ascent is equal to the time of descent.

(C) When objects are dropped from a height,they undergo free fall under the influence of gravity.
In the absence of air resistance,the acceleration of an object in free fall is equal to the acceleration due to gravity $(g)$,which is approximately $9.8 \, m/s^2$.
This acceleration is independent of the mass,size,or material of the object.
Since both spheres are dropped simultaneously from the same height,at any given point (including $10 \, m$ above the ground),they will have the same velocity and the same acceleration $(g)$.
Potential energy $(PE = mgh)$ depends on mass,so it will be different for the two spheres.
Momentum $(p = mv)$ depends on mass,so it will be different.
Kinetic energy $(KE = 1/2 mv^2)$ depends on mass,so it will be different.
Therefore,the only quantity that remains the same for both spheres is their acceleration.