Free Fall and Acceleration due to Gravity Questions in English

(N/A) For upward motion:
$u = u, v = 0$
Let $t_{1}$ be the time of ascent. Using the equation of motion $v = u + at$:
$0 = u - gt_{1}$
$t_{1} = u/g$
For downward motion (descent):
The body starts from rest at the maximum height,so $u = 0$. Let $t_{2}$ be the time of descent. Using $s = ut + (1/2)at^{2}$ where $s = h$:
$h = 0 + (1/2)gt_{2}^{2} \implies t_{2} = \sqrt{2h/g}$
Since $h = u^{2}/(2g)$,substituting this gives $t_{2} = \sqrt{2(u^{2}/2g)/g} = u/g$.
Thus,$t_{1} = t_{2}$,which proves that the time of ascent is equal to the time of descent.

(A) Let stone $A$ fall from the top of a tower of height $h = 100 \, m$. Its initial velocity $u_1 = 0$ and acceleration $a_1 = +g = 10 \, m s^{-2}$.
Another stone $B$ is projected upwards from the ground with initial velocity $u_2 = 25 \, m s^{-1}$ and acceleration $a_2 = -g = -10 \, m s^{-2}$.
Let the two stones meet at a point $C$ at a distance $y$ below the top of the tower after time $t$. For stone $A$:
$y = u_1 t + \frac{1}{2} a_1 t^2 = 0 + \frac{1}{2} \times 10 \times t^2 = 5t^2$ ... $(1)$
For stone $B$,the distance covered from the ground is $(100 - y)$:
$100 - y = u_2 t + \frac{1}{2} a_2 t^2 = 25t - \frac{1}{2} \times 10 \times t^2 = 25t - 5t^2$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$y + (100 - y) = 5t^2 + 25t - 5t^2$
$100 = 25t$
$t = 4 \, s$
Substituting $t = 4 \, s$ into equation $(1)$:
$y = 5 \times (4)^2 = 5 \times 16 = 80 \, m$.
The stones will meet after $4 \, s$ at a distance of $80 \, m$ from the top of the tower (or $20 \, m$ from the ground).

(B) For stone $A$ dropped from $h_1 = 100 \, m$:
Using the equation $v^2 - u^2 = 2gh$,where initial velocity $u = 0$:
$v_A^2 = 2 \times 10 \times 100 = 2000$
$v_A = \sqrt{2000} \approx 44.72 \, m s^{-1}$.
For stone $B$ dropped from $h_2 = 50 \, m$:
Using the equation $v^2 - u^2 = 2gh$,where initial velocity $u = 0$:
$v_B^2 = 2 \times 10 \times 50 = 1000$
$v_B = \sqrt{1000} \approx 31.62 \, m s^{-1}$.
Since $v_A \neq v_B$,the stones will not have equal velocities upon reaching the ground. Stone $A$ will have a higher velocity.

(N/A) Newton's universal law of gravitation states that 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 their centers.
Mathematically,the force $F$ is given by:
$F = \frac{G m_1 m_2}{r^2}$
where $G$ is the universal gravitational constant,$m_1$ and $m_2$ are the masses of the two objects,and $r$ is the distance between their centers.
To show that acceleration due to gravity $(g)$ is independent of the mass of the falling object,consider an object of mass $m$ falling towards the Earth (mass $M$,radius $R$).
The gravitational force is $F = \frac{G M m}{R^2}$.
According to Newton's second law,$F = m \times a$. Here,$a = g$,so $F = m \times g$.
Equating the two expressions: $m \times g = \frac{G M m}{R^2}$.
Canceling $m$ from both sides,we get $g = \frac{G M}{R^2}$.
Since $g$ depends only on the mass of the Earth $(M)$,the gravitational constant $(G)$,and the radius of the Earth $(R)$,it is independent of the mass $(m)$ of the falling object.

(N/A) Following are the three differences:
$(i)$ The value of $g$ is $9.8 \, m/s^2$ on the surface of the Earth,whereas the value of $G$ is $6.67 \times 10^{-11} \, N \cdot m^2/kg^2$.
$(ii)$ The value of $g$ varies from place to place,whereas the value of $G$ remains constant throughout the universe.
$(iii)$ $g$ is a vector quantity,while $G$ is a scalar quantity.

(N/A) Acceleration due to gravity is the acceleration produced in a body when it is allowed to fall freely under the influence of gravity alone. The acceleration due to gravity $(g)$ on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $M$ is the mass of the planet and $R$ is its radius.
Given:
Mass of the planet $(M_P)$ = $\frac{M_E}{2}$
Radius of the planet $(R_P)$ = $\frac{R_E}{2}$
Acceleration due to gravity on Earth $(g_E)$ = $9.8 \, m/s^2$
Using the ratio:
$\frac{g_P}{g_E} = \frac{GM_P / R_P^2}{GM_E / R_E^2} = \frac{M_P}{M_E} \times (\frac{R_E}{R_P})^2$
Substituting the values:
$\frac{g_P}{g_E} = \frac{M_E / 2}{M_E} \times (\frac{R_E}{R_E / 2})^2 = \frac{1}{2} \times (2)^2 = \frac{1}{2} \times 4 = 2$
Therefore,$g_P = 2 \times g_E = 2 \times 9.8 \, m/s^2 = 19.6 \, m/s^2$.

(N/A) The acceleration due to gravity is given by the expression $g = \frac{GM}{R^2}$.
This depends upon the radius of the Earth $(R)$. Since the Earth is not a perfect sphere and its radius varies from the poles to the equator,the value of $g$ is not constant everywhere. Additionally,$g$ varies with altitude and depth.
$(b)$ The acceleration due to gravity $(g)$ acting on a falling body is given by $g = \frac{GM}{R^2}$.
This expression shows that $g$ does not depend on the mass of the falling object $(m)$,but only on the mass of the planet $(M)$ and the distance from its center $(R)$. Therefore,all objects fall with the same acceleration,regardless of their mass,and take the same time to fall from a fixed height in a vacuum.
$(c)$ '$G$' is known as the universal gravitational constant because its value remains the same at all places in the universe and is independent of the nature of the medium or the bodies involved.

(N/A) Free fall is the motion of an object falling solely under the influence of gravitational force,with constant acceleration.
$(b)$ The direction of motion remains the same (vertically downwards) during free fall.
$(c)$ Given: Initial velocity $u = 0 \, m/s$,height $h = 19.6 \, m$,acceleration $g = 9.8 \, m/s^2$.
$(i)$ Using the equation of motion $h = ut + \frac{1}{2}gt^2$:
$19.6 = 0 + \frac{1}{2} \times 9.8 \times t^2$
$19.6 = 4.9 \times t^2$
$t^2 = 4$
$t = 2 \, s$.
$(ii)$ Using the equation $v = u + gt$:
$v = 0 + 9.8 \times 2$
$v = 19.6 \, m/s$.
$(iii)$ Since the object is in free fall,the acceleration is constant and equal to the acceleration due to gravity,which is $9.8 \, m/s^2$ at any time,including after $1 \, s$.

(A) Given:
Height of the object,$h = 10 \, m$
Final velocity at the highest point,$v = 0 \, m/s$
Acceleration due to gravity,$a = -g = -9.8 \, m/s^2$
$(i)$ Using the third equation of motion,$v^2 - u^2 = 2ah$:
$0^2 - u^2 = 2 \times (-9.8) \times 10$
$-u^2 = -196$
$u^2 = 196$
$u = 14 \, m/s$
Thus,the initial velocity is $14 \, m/s$.
$(ii)$ Using the first equation of motion,$v = u + at$:
$0 = 14 + (-9.8) \times t$
$9.8t = 14$
$t = 14 / 9.8 \approx 1.43 \, s$
Thus,the time taken to reach the highest point is approximately $1.43 \, s$.

(B) We know that the acceleration due to gravity on Earth is given by $g_{e} = \frac{GM_{e}}{R_{e}^{2}}$.
For the given planet,the mass $M_{p} = \frac{M_{e}}{2}$ and the radius $R_{p} = \frac{R_{e}}{2}$.
Substituting these values into the formula for the planet's gravity $g_{p} = \frac{GM_{p}}{R_{p}^{2}}$,we get:
$g_{p} = \frac{G(M_{e}/2)}{(R_{e}/2)^{2}} = \frac{GM_{e}/2}{R_{e}^{2}/4} = \frac{GM_{e}}{2} \times \frac{4}{R_{e}^{2}} = 2 \times \frac{GM_{e}}{R_{e}^{2}}$.
Since $\frac{GM_{e}}{R_{e}^{2}} = g_{e}$,it follows that $g_{p} = 2g_{e}$.

(B) When an object falls freely under the action of gravity,it experiences a constant acceleration due to gravity $(g \approx 9.8 \ m/s^2)$.
Since the acceleration remains constant throughout the fall,the velocity of the object increases by an equal amount in equal intervals of time.
Therefore,the nature of the motion is uniformly accelerated motion in one dimension.

(B) The acceleration due to gravity $(g)$ on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the universal gravitational constant,$M$ is the mass of the planet,and $R$ is the radius of the planet.
From this relation,it is clear that $g \propto M$.
Therefore,the acceleration due to gravity is directly proportional to the mass of the planet.

(A) Acceleration due to gravity decreases with increasing altitude because $g' = g(1 - 2h/R_e)$ for small heights.
$(b)$ Acceleration due to gravity decreases with increasing depth because $g' = g(1 - d/R_e)$.
$(c)$ Acceleration due to gravity is independent of the mass of the body,as $g = GM/R^2$,where $M$ is the mass of the Earth and $R$ is its radius.

(B) The acceleration due to gravity on the surface of the Moon is approximately $1/6$th of that on the surface of the Earth.
Since the gravitational force is much weaker on the Moon,a person can jump $6$ times higher on the Moon compared to the Earth for the same amount of muscular effort.

(A) $1$. First,calculate the mass of the body $(m)$. Given weight $W = mg = 63 \, N$. Assuming $g = 10 \, m/s^2$,$m = 63 / 10 = 6.3 \, kg$.
$2$. The distance of the body from the center of the earth $(r)$ at a height $h = R/2$ is $r = R + h = R + R/2 = 1.5R$.
$3$. Given $R = 6.4 \times 10^6 \, m$,so $r = 1.5 \times 6.4 \times 10^6 = 9.6 \times 10^6 \, m$.
$4$. The gravitational force is given by $F = \frac{GMm}{r^2}$.
$5$. Substituting the values: $F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 6.3}{(9.6 \times 10^6)^2}$.
$6$. $F = \frac{252.186 \times 10^{13}}{92.16 \times 10^{12}} = \frac{2521.86}{92.16} \approx 27.35 \, N$.

The formula for acceleration due to gravity $(g)$ is given by $g = \frac{GM}{R^{2}}$,where $G$ is the universal gravitational constant,$M$ is the mass of the celestial body,and $R$ is its radius.
Rearranging the formula to solve for mass $(M)$,we get $M = \frac{g R^{2}}{G}$.
Given values are:
$g = 1.67 \, m s^{-2}$
$R = 1.74 \times 10^{6} \, m$
$G = 6.67 \times 10^{-11} \, N m^{2} kg^{-2}$
Substituting these values into the formula:
$M = \frac{1.67 \times (1.74 \times 10^{6})^{2}}{6.67 \times 10^{-11}}$
$M = \frac{1.67 \times 3.0276 \times 10^{12}}{6.67 \times 10^{-11}}$
$M \approx 0.758 \times 10^{23} \, kg = 7.58 \times 10^{22} \, kg$.
Thus,the mass of the moon is approximately $7.58 \times 10^{22} \, kg$.

(N/A) Given: Initial velocity,$u = 0 \, m/s$.
Acceleration due to gravity,$g = 9.8 \, m/s^2$.
$(a)$ Using the equation of motion $S = ut + \frac{1}{2}at^2$:
$4.9 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$4.9 = 4.9 \times t^2$
$t^2 = 1 \Rightarrow t = 1 \, s$.
So,it takes $1 \, s$ to fall $4.9 \, m$.
$(b)$ Using $v^2 - u^2 = 2aS$:
$v^2 - 0^2 = 2 \times 9.8 \times 4.9$
$v^2 = 96.04$
$v = 9.8 \, m/s$.
So,the speed at the end of $4.9 \, m$ is $9.8 \, m/s$.
$(c)$ Using $v^2 - u^2 = 2aS$:
$v^2 - 0^2 = 2 \times 9.8 \times 7.9$
$v^2 = 154.84$
$v = \sqrt{154.84} \approx 12.44 \, m/s$.
So,the speed at the end of $7.9 \, m$ is $12.44 \, m/s$.
$(d)$ During free fall,the acceleration due to gravity is constant.
Acceleration after $1 \, s = 9.8 \, m/s^2$.
Acceleration after $2 \, s = 9.8 \, m/s^2$.

(12.1 M/S) For the first stone,we are given:
$u = 0\, m\, s^{-1}, h = 49\, m, g = 9.8\, m\, s^{-2}$
Using the equation of motion $S = ut + \frac{1}{2}at^2$,we have:
$49 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$t^2 = \frac{98}{9.8} = 10$
$t = \sqrt{10} \approx 3.16\, s$
So,the first stone takes $3.16\, s$ to reach the ground.
For the second stone,it is thrown $1\, s$ later but hits the ground at the same time. Therefore,the time taken by the second stone is $t_2 = 3.16 - 1 = 2.16\, s$.
Using $S = ut + \frac{1}{2}at^2$ for the second stone:
$49 = u \times 2.16 + \frac{1}{2} \times 9.8 \times (2.16)^2$
$49 = 2.16u + 4.9 \times 4.6656$
$49 = 2.16u + 22.86$
$2.16u = 49 - 22.86 = 26.14$
$u = \frac{26.14}{2.16} \approx 12.10\, m\, s^{-1}$
Thus,the second stone was thrown with a speed of $12.1\, m\, s^{-1}$.

(19.42 M) Let the distance between the top of the window and the roof be $S$. For the journey across the window,i.e.,from $B$ to $C$:
Let the velocity at $B = u \, m/s$.
Distance travelled,$S = 2 \, m$.
Time taken,$t = 0.1 \, s$.
Acceleration,$a = g = 9.8 \, m/s^2$.
Using the equation $S = ut + \frac{1}{2}at^2$,we have:
$2 = u(0.1) + \frac{1}{2}(9.8)(0.1)^2$
$2 = 0.1u + 0.049$
$0.1u = 1.951$
$u = 19.51 \, m/s$.
Therefore,the velocity of the stone at the top of the window is $19.51 \, m/s$.
Now,for the free fall of the stone from the roof top to the top of the window,i.e.,from point $A$ to $B$:
Initial velocity $u_i = 0 \, m/s$.
Final velocity $v = 19.51 \, m/s$.
Distance covered $S = ?$.
Acceleration due to gravity $g = 9.8 \, m/s^2$.
Using $v^2 - u_i^2 = 2gS$,we have:
$(19.51)^2 - 0^2 = 2 \times 9.8 \times S$
$S = \frac{(19.51)^2}{19.6} \approx 19.42 \, m$.

(A) For a ball thrown vertically upwards:
Initial velocity $= u \text{ m/s}$.
Final velocity at maximum height $= v = 0 \text{ m/s}$.
Using the equation of motion $v = u + at$,where $a = -g$ (acceleration due to gravity acting downwards):
$0 = u - gt \implies t = u/g$ (Time taken to reach maximum height).
For the return journey,the ball starts from rest at the maximum height $(u_{return} = 0)$ and falls downwards under gravity $(a = +g)$:
Using the equation $v^2 = u^2 + 2as$:
$v^2 = 0^2 + 2gH$,where $H$ is the maximum height.
From the upward journey,$H = u^2 / (2g)$.
Substituting $H$ in the return equation: $v^2 = 2g(u^2 / 2g) = u^2$.
Therefore,$v = u$.
The ball returns to the point of projection with the same velocity $u$ with which it was thrown.

(78.4 M) Given: Initial velocity $u = 0 \, m \, s^{-1}$,time $t = 4 \, s$,and acceleration due to gravity $g = 9.8 \, m \, s^{-2}$.
Using the second equation of motion for free fall:
$h = ut + \frac{1}{2}gt^2$
Substituting the values:
$h = (0 \times 4) + \frac{1}{2} \times 9.8 \times (4)^2$
$h = 0 + 0.5 \times 9.8 \times 16$
$h = 4.9 \times 16$
$h = 78.4 \, m$
Thus,the height of the building is $78.4 \, m$.

(N/A) Given: Total time $T = 6 \, s$,final velocity at maximum height $v = 0 \, m s^{-1}$,acceleration due to gravity $g = 9.8 \, m s^{-2}$.
Since the time of ascent equals the time of descent,the time taken to reach the maximum height is $t = T / 2 = 6 / 2 = 3 \, s$.
$(a)$ Using the first equation of motion,$v = u + at$,where $a = -g$:
$0 = u + (-9.8 \times 3)$
$u = 29.4 \, m s^{-1}$.
Thus,the initial velocity is $29.4 \, m s^{-1}$.
$(b)$ Using the third equation of motion,$v^2 - u^2 = 2as$:
$0^2 - (29.4)^2 = 2 \times (-9.8) \times h$
$-864.36 = -19.6 \times h$
$h = 864.36 / 19.6 = 44.1 \, m$.
Thus,the maximum height attained is $44.1 \, m$.

(N/A) Given: height $h = 20\, m$,initial velocity $u = 0$,acceleration due to gravity $g = 10\, m s^{-2}$.
Using the second equation of motion $S = ut + \frac{1}{2}at^2$,where $S = h$ and $a = g$:
$20 = 0 \times t + \frac{1}{2} \times 10 \times t^2$
$20 = 5t^2$
$t^2 = 4$
$t = 2\, s$.
So,it will take $2\, s$ to reach the ground.
$(b)$ Using the first equation of motion $v = u + at$:
$v = 0 + 10 \times 2$
$v = 20\, m s^{-1}$.
Thus,the speed when it hits the ground is $20\, m s^{-1}$.

(D) The acceleration due to gravity $g$ 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.
Because the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius $R$ is smallest at the poles and largest at the equator.
Since $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,the value of $g$ is maximum where the radius $R$ is minimum.
The radius of the Earth is minimum at the poles (Antarctica is located near the South Pole).
Therefore,the value of $g$ is maximum at a camp site in Antarctica.

(B) When a ball is thrown vertically upward,its velocity is directed upwards.
However,the acceleration due to gravity $(g)$ always acts vertically downwards towards the center of the Earth.
Since the motion of the ball is upwards and the acceleration due to gravity is downwards,the acceleration is in the direction opposite to the direction of motion.
Therefore,the correct option is $B$.

(C) When a projectile is at the highest point of its trajectory,its vertical velocity becomes $0 \ m/s$,but it still possesses a horizontal velocity component.
However,the acceleration due to gravity $(g)$ acts constantly on the projectile throughout its entire flight.
This acceleration is always directed vertically downwards towards the center of the Earth.
Therefore,even at the top of its path,the projectile experiences a constant downward acceleration equal to $g$.

(D) The acceleration due to gravity $(g)$ at a height $(h)$ above the Earth's surface is given by $g_h = g(1 - 2h/R)$,where $R$ is the radius of the Earth. As $h$ increases,$g_h$ decreases.
Similarly,the acceleration due to gravity at a depth $(d)$ below the Earth's surface is given by $g_d = g(1 - d/R)$. As $d$ increases,$g_d$ decreases.
Therefore,for both scientist $A$ (moving deep into a mine) and scientist $B$ (moving high up in a balloon),the measured acceleration due to gravity decreases.

(B) The effective acceleration due to gravity '$g'$ at the equator is given by the formula: $g' = g - R\omega^2$,where '$g$' is the acceleration due to gravity at the poles (or if the earth were stationary),'$R$' is the radius of the earth,and '$\omega$' is the angular velocity of the earth's rotation.
If the earth stops rotating,the angular velocity '$\omega$' becomes $0$.
Substituting this into the formula,we get: $g' = g - R(0)^2 = g$.
Since the original value was $g' = g - R\omega^2$ (which is less than '$g$'),the new value '$g$' is greater than the original value '$g'$ at the equator.
Therefore,the value of '$g$' at the equator will increase.

(B) The acceleration due to gravity on the surface of the Earth is defined as the standard value $g$.
By definition,the gravitational acceleration experienced by an object at the Earth's surface is denoted by the symbol $g$,which is approximately $9.8 \ m/s^2$.
Therefore,the correct option is $B$.

(B) In a vacuum,there is no air resistance to oppose the motion of a falling object.
According to Newton's second law of motion,$F = ma$,and the gravitational force acting on an object is $F = mg$,where $m$ is the mass of the object and $g$ is the acceleration due to gravity.
Equating these,$ma = mg$,which simplifies to $a = g$.
Since $g$ is a constant value (approximately $9.8 \ m/s^2$ near the Earth's surface) and is independent of the mass of the object,all objects falling freely in a vacuum experience the same acceleration.

(C) The gravitational acceleration $g$ at a distance $r$ from the centre of the Earth is given by the formula $g' = \frac{GM}{r^2}$.
At the surface of the Earth,the distance $r = R$ (where $R$ is the radius of the Earth),so $g = \frac{GM}{R^2} = 9.8\, m s^{-2}$.
Given that the spaceship is at a distance $r = 2R$ from the centre of the Earth.
Substituting this into the formula: $g' = \frac{GM}{(2R)^2} = \frac{GM}{4R^2}$.
Since $g = \frac{GM}{R^2}$,we can write $g' = \frac{g}{4}$.
Substituting the value of $g = 9.8\, m s^{-2}$,we get $g' = \frac{9.8}{4} = 2.45\, m s^{-2}$.

(D) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.
For Earth,$g_e = \frac{GM_e}{R_e^2} = 9.8 \, m s^{-2}$.
For the new planet,the mass $M_p = \frac{M_e}{2}$ and the radius $R_p = \frac{R_e}{2}$.
Substituting these values into the formula for the new planet's gravity $g_p$:
$g_p = \frac{G(M_e/2)}{(R_e/2)^2} = \frac{G M_e / 2}{R_e^2 / 4} = 2 \times \frac{GM_e}{R_e^2}$.
Since $\frac{GM_e}{R_e^2} = g_e$,we have $g_p = 2 \times g_e$.
$g_p = 2 \times 9.8 \, m s^{-2} = 19.6 \, m s^{-2}$.

(A) Given:
Initial velocity $u = 0 \, m s^{-1}$ (since the stone is dropped).
Distance covered $s = 100 \, m$.
Acceleration due to gravity $g = 9.8 \, m s^{-2}$.
Using the third equation of motion: $v^2 - u^2 = 2as$.
Substituting the values: $v^2 - 0^2 = 2 \times 9.8 \times 100$.
$v^2 = 1960$.
$v = \sqrt{1960} \approx 44.27 \, m s^{-1}$.
Thus,the speed is approximately $44.2 \, m s^{-1}$.

(C) Given:
Initial velocity $(u)$ = $0\, m/s$ (since the stone is dropped).
Time $(t)$ = $4\, s$.
Acceleration due to gravity $(g)$ = $9.8\, m/s^2$.
Using the second equation of motion:
$s = ut + \frac{1}{2}gt^2$
Substituting the values:
$h = (0 \times 4) + \frac{1}{2} \times 9.8 \times (4)^2$
$h = 0 + 4.9 \times 16$
$h = 78.4\, m$
Therefore,the height is $78.4\, m$.

(A) The initial velocity $u = 0 \, m/s$ and the acceleration due to gravity is $g$.
Using the equation of motion $s = ut + (1/2)gt^2$,for the total height $H = 100 \, m$ and time $T$:
$100 = 0 \cdot T + (1/2)gT^2 \implies 100 = (1/2)gT^2 \implies T^2 = 200/g$.
Now,let $h$ be the distance covered in time $t = T/2$:
$h = (1/2)g(T/2)^2 = (1/2)g(T^2/4) = (1/8)gT^2$.
Substituting $gT^2 = 200$ into the equation:
$h = (1/8) \cdot 200 = 25 \, m$.
Since $h$ is the distance covered from the top,the bottle will be $25 \, m$ from the top.

(A) Let the two bodies be $A$ and $B$. Body $A$ is $1 \, m$ above body $B$.
Both bodies are released simultaneously and fall freely under gravity,meaning their initial velocity $u = 0 \, m/s$ and acceleration $a = g = 9.8 \, m/s^2$ are the same for both.
The position of body $A$ at time $t$ is given by $y_A = y_{A,0} + ut + \frac{1}{2}gt^2$.
The position of body $B$ at time $t$ is given by $y_B = y_{B,0} + ut + \frac{1}{2}gt^2$.
The relative separation between the bodies at any time $t$ is given by $|y_A - y_B| = |(y_{A,0} + ut + \frac{1}{2}gt^2) - (y_{B,0} + ut + \frac{1}{2}gt^2)|$.
This simplifies to $|y_A - y_B| = |y_{A,0} - y_{B,0}|$.
Since the initial vertical separation $|y_{A,0} - y_{B,0}|$ is $1 \, m$,the relative separation remains constant at $1 \, m$ regardless of the time elapsed,as both bodies experience the same acceleration due to gravity.

(TRUE) True. When a spring balance holding an object is released,it falls freely under the influence of gravity. In a state of free fall,both the spring balance and the object experience the same acceleration due to gravity $(g)$. Consequently,there is no net force exerted by the object on the spring of the balance,leading to a reading of zero weight.

(TRUE) True.
When a body is thrown vertically upward,it experiences a constant downward acceleration due to gravity $(g)$,which is approximately $9.8 \ m/s^2$.
When a body falls downward,it also experiences the same constant downward acceleration due to gravity $(g = 9.8 \ m/s^2)$.
In physics,if we take the upward direction as positive $(+)$,then the downward acceleration is represented as negative $(-g)$.
Therefore,the magnitude remains the same $(9.8 \ m/s^2)$,but the sign is opposite depending on the chosen coordinate system.

(TRUE) The statement is True.
According to the formula for acceleration due to gravity at a depth $d$ below the surface of the Earth,$g_d = g(1 - d/R)$,where $g$ is the acceleration due to gravity at the surface,$d$ is the depth,and $R$ is the radius of the Earth.
At the centre of the Earth,the depth $d$ is equal to the radius $R$ $(d = R)$.
Substituting $d = R$ into the formula,we get $g_d = g(1 - R/R) = g(1 - 1) = g(0) = 0$.
Therefore,the acceleration due to gravity $(g)$ at the centre of the Earth is zero.

(TRUE) True.
The acceleration due to gravity $g$ on the surface of a planet is defined by the formula $g = GM/R^2$,where $G$ is the universal gravitational constant,$M$ is the mass of the planet,and $R$ is the radius of the planet.

(B) Given: Mass $(m) = 10 \, kg$,Initial velocity $(u) = 0 \, m s^{-1}$,Height $(S) = 0.8 \, m$,Acceleration due to gravity $(g) = 10 \, m s^{-2}$.
Using the equation of motion $v^2 - u^2 = 2gS$:
$v^2 - (0)^2 = 2 \times 10 \times 0.8$
$v^2 = 16$
$v = 4 \, m s^{-1}$.
Momentum $(p)$ is defined as the product of mass and velocity:
$p = m \times v$
$p = 10 \, kg \times 4 \, m s^{-1} = 40 \, kg \, m s^{-1}$ (or $40 \, N s$).
Thus,the dumbbell will transfer $40 \, kg \, m s^{-1}$ of momentum to the floor.