Motion Under Gravity Questions in English

(A) The correct statement is $(A)$.
When a body is projected vertically upward,at the highest point of its trajectory,its velocity becomes $0 \ m/s$.
However,the acceleration due to gravity acting on the body remains $g = 9.8 \ m/s^2$ downwards.
Therefore,a body with zero velocity does not necessarily have zero acceleration.

(B) $1$. During the fall,the tennis ball moves downward,so its displacement and velocity are directed downward.
$2$. Upon hitting the ground and bouncing back,the ball moves upward,so the directions of both displacement and velocity change to upward.
$3$. Throughout the entire motion (falling and rising),the acceleration due to gravity $(g)$ always acts vertically downward.
$4$. Therefore,only the directions of displacement and velocity change,while the direction of acceleration remains constant.

(C) Given: Initial velocity of the balloon (and thus the stone) $u = 12 \, m/s$ (downward).
Acceleration due to gravity $g = 9.8 \, m/s^2$ (downward).
Time $t = 10 \, s$.
Since the stone is released from a descending balloon,its initial velocity is in the direction of motion (downward).
Taking the downward direction as positive,the displacement $s$ is given by the kinematic equation:
$s = ut + \frac{1}{2}gt^2$
Substituting the values:
$s = (12 \, m/s)(10 \, s) + \frac{1}{2}(9.8 \, m/s^2)(10 \, s)^2$
$s = 120 + 0.5 \times 9.8 \times 100$
$s = 120 + 490$
$s = 610 \, m$.
Therefore,the displacement of the stone after $10 \, s$ is $610 \, m$.

(B) Velocity at the time of striking the floor $(u_1)$:
$u_1 = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \, m/s$ (downwards).
Taking downward direction as negative,$u_1 = -14 \, m/s$.
Velocity with which it rebounds $(v_1)$:
$v_1 = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 2.5} = \sqrt{49} = 7 \, m/s$ (upwards).
Taking upward direction as positive,$v_1 = +7 \, m/s$.
Change in velocity $\Delta v = v_1 - u_1 = 7 - (-14) = 21 \, m/s$.
Average acceleration $a = \frac{\Delta v}{\Delta t} = \frac{21}{0.01} = 2100 \, m/s^2$.
Since the change in velocity is positive,the acceleration is directed upwards.

(D) Let $t$ be the time elapsed since body $A$ was projected. The displacement of body $A$ is given by $h_A = ut - \frac{1}{2}gt^2$.
Since body $B$ is projected $4 \, s$ later,its time of flight is $(t-4) \, s$. Its displacement is $h_B = u(t-4) - \frac{1}{2}g(t-4)^2$.
When the bodies meet,$h_A = h_B$.
$98t - \frac{1}{2} \times 9.8 \times t^2 = 98(t-4) - \frac{1}{2} \times 9.8 \times (t-4)^2$.
$98t - 4.9t^2 = 98t - 392 - 4.9(t^2 - 8t + 16)$.
$98t - 4.9t^2 = 98t - 392 - 4.9t^2 + 39.2t - 78.4$.
$0 = -392 + 39.2t - 78.4$.
$39.2t = 470.4$.
$t = \frac{470.4}{39.2} = 12 \, s$.

(C) For a body dropped from rest,the distance $h$ covered in time $t$ is given by the equation of motion: $h = \frac{1}{2}gt^2$.
Rearranging for time $t$,we get: $t = \sqrt{\frac{2h}{g}}$.
For the first body dropped from height $a$,the time taken is $t_a = \sqrt{\frac{2a}{g}}$.
For the second body dropped from height $b$,the time taken is $t_b = \sqrt{\frac{2b}{g}}$.
The ratio of the time taken is $\frac{t_a}{t_b} = \frac{\sqrt{2a/g}}{\sqrt{2b/g}} = \sqrt{\frac{a}{b}}$.
Thus,the ratio is $\sqrt{a} : \sqrt{b}$.

(B) Let the total time of motion be $n$ seconds.
Distance covered in the first $3$ seconds is given by $s_3 = \frac{1}{2} g (3)^2 = 4.5g$.
Distance covered in the last second of motion is given by $s_{last} = u + \frac{g}{2} (2n - 1)$.
Since the body falls from rest,$u = 0$,so $s_{last} = \frac{g}{2} (2n - 1)$.
According to the problem,$s_{last} = s_3$.
Therefore,$\frac{g}{2} (2n - 1) = \frac{1}{2} g (3)^2$.
Canceling $\frac{g}{2}$ from both sides,we get $2n - 1 = 9$.
$2n = 10$,which gives $n = 5 \; s$.

(A) Let $t$ be the time taken by the first stone to reach the water surface. Using the equation of motion $h = ut + \frac{1}{2}gt^2$,where $u = 0$ (dropped),$h = 44.1 \ m$,and $g = 9.8 \ m/s^2$:
$44.1 = 0 + \frac{1}{2} \times 9.8 \times t^2$
$t^2 = \frac{44.1 \times 2}{9.8} = 9$
$t = 3 \ s$
Since the second stone is thrown $1 \ s$ later and reaches the water at the same time,the time taken by the second stone is $t' = 3 - 1 = 2 \ s$.
Using the equation $h = u't' + \frac{1}{2}g(t')^2$ for the second stone:
$44.1 = u' \times 2 + \frac{1}{2} \times 9.8 \times (2)^2$
$44.1 = 2u' + 19.6$
$2u' = 44.1 - 19.6 = 24.5$
$u' = 12.25 \ m/s$.

(B) Let the initial velocity of the body be $u$ and the acceleration due to air resistance be $a$.
During the upward motion,both gravity $(g)$ and air resistance $(a)$ act downwards. The net acceleration is $(g + a)$. The time of rise $t_1$ is given by $t_1 = \frac{u}{g + a}$.
The maximum height reached is $H = \frac{u^2}{2(g + a)}$.
During the downward motion,gravity acts downwards while air resistance acts upwards. The net acceleration is $(g - a)$. The time of fall $t_2$ is given by $H = \frac{1}{2}(g - a)t_2^2$.
Substituting $H$,we get $\frac{1}{2}(g - a)t_2^2 = \frac{u^2}{2(g + a)}$,which simplifies to $t_2 = \frac{u}{\sqrt{(g + a)(g - a)}}$.
Comparing $t_1$ and $t_2$,we see that $t_2 = t_1 \sqrt{\frac{g + a}{g - a}}$. Since $\sqrt{\frac{g + a}{g - a}} > 1$,it follows that $t_2 > t_1$. Thus,the time of rise is less than the time of fall.

(D) Let the first body be released at $t = 0$. Its position after $t$ seconds is given by $y_1 = \frac{1}{2}gt^2$.
The second body is released at $t = 1 \; s$. Its position after $t$ seconds (where $t > 1$) is given by $y_2 = \frac{1}{2}g(t - 1)^2$.
We need to find the separation $2 \; s$ after the release of the second body. This corresponds to a total time of $t = 1 + 2 = 3 \; s$ from the release of the first body.
Position of the first body at $t = 3 \; s$: $y_1 = \frac{1}{2} \times 9.8 \times (3)^2 = 4.9 \times 9 = 44.1 \; m$.
Position of the second body at $t = 3 \; s$ (which is $2 \; s$ after its release): $y_2 = \frac{1}{2} \times 9.8 \times (2)^2 = 4.9 \times 4 = 19.6 \; m$.
Separation between the two bodies: $\Delta y = y_1 - y_2 = 44.1 - 19.6 = 24.5 \; m$.

(B) The time of flight $T$ for an object projected vertically upwards with initial velocity $u$ is given by the formula $T = \frac{2u}{g}$.
Given,initial velocity $u = 100 \, m/s$ and acceleration due to gravity $g \approx 10 \, m/s^2$.
Substituting the values into the formula:
$T = \frac{2 \times 100}{10} = \frac{200}{10} = 20 \, sec$.
Therefore,the object will strike the ground after $20 \, sec$.

(A) Given that the stone is dropped from the top of the tower,its initial velocity $u = 0 \, m/s$.
The time taken to reach the ground is $t = 4 \, s$.
Taking the acceleration due to gravity $g = 10 \, m/s^2$.
Using the second equation of motion for vertical distance $h$:
$h = ut + \frac{1}{2}gt^2$
Substituting the values:
$h = 0 \times 4 + \frac{1}{2} \times 10 \times (4)^2$
$h = 0 + 5 \times 16$
$h = 80 \, m$.
Therefore,the height of the tower is $80 \, m$.

(D) Let the initial position be at the top of the tower $(y = h)$ and the ground be at $y = 0$. The body is released,so initial velocity $u = 0$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,the total height $h$ is given by $h = \frac{1}{2}gt^2$ ... $(i)$
After time $t/2$,the distance covered from the top is $x = \frac{1}{2}g(t/2)^2 = \frac{1}{2}g(t^2/4) = \frac{1}{8}gt^2$ ... (ii)
From equation $(i)$,we know $\frac{1}{2}gt^2 = h$,so substituting this into equation (ii),we get $x = h/4$.
This $x$ represents the distance from the top of the tower.
The height from the ground is $h - x = h - h/4 = 3h/4$.

(A) The equation of motion for a particle under gravity is given by $h = ut - \frac{1}{2}gt^2$.
Here,$u = 80 \; ft/sec$,$h = 96 \; ft$,and $g = 32 \; ft/sec^2$.
Substituting these values into the equation:
$96 = 80t - \frac{1}{2}(32)t^2$
$96 = 80t - 16t^2$
Dividing the entire equation by $16$:
$6 = 5t - t^2$
Rearranging into a standard quadratic form:
$t^2 - 5t + 6 = 0$
Factoring the quadratic equation:
$(t - 2)(t - 3) = 0$
Thus,$t = 2 \; sec$ or $t = 3 \; sec$.
The particle reaches a height of $96 \; ft$ at $2 \; sec$ while moving upwards and at $3 \; sec$ while moving downwards.

(B) Given that the body falls from rest,the initial velocity $u = 0 \ ft/sec$.
The acceleration due to gravity is $g = 32 \ ft/sec^2$.
The time elapsed is $t = 1 \ sec$.
Using the first equation of motion for a body falling under gravity:
$v = u + gt$
Substituting the values:
$v = 0 + (32 \ ft/sec^2) \times (1 \ sec)$
$v = 32 \ ft/sec$.
Therefore,the velocity at the end of the first second is $32 \ ft/sec$.

(B) Let the upward direction be positive and the downward direction be negative.
Initial velocity $u_i = +u$.
Final velocity at the ground $v_f = -3u$.
Acceleration due to gravity $a = -g$.
Let the height of the tower be $h$. The displacement $s = -h$.
Using the third equation of motion: $v_f^2 = u_i^2 + 2as$.
Substituting the values: $(-3u)^2 = (u)^2 + 2(-g)(-h)$.
$9u^2 = u^2 + 2gh$.
$8u^2 = 2gh$.
$h = 4u^2/g$.

(C) When an object is dropped from a height $h$ under gravity,its motion is governed by the equation of motion $h = ut + \frac{1}{2}gt^2$. Since the initial velocity $u = 0$,the equation simplifies to $h = \frac{1}{2}gt^2$. Solving for time $t$,we get $t = \sqrt{\frac{2h}{g}}$. As the time of flight $t$ depends only on the height $h$ and the acceleration due to gravity $g$,and is independent of the mass of the object,both stones will reach the ground simultaneously.

(B) The body is thrown vertically upwards and returns to the ground. The total time of flight $T$ is given by the formula $T = \frac{2u}{g}$.
Given initial speed $u = 96\,ft/\sec$ and acceleration due to gravity $g = 32\,ft/\sec^2$.
Substituting the values: $T = \frac{2 \times 96}{32} = \frac{192}{32} = 6\,sec$.
Therefore,the body reaches the ground after $6\,sec$.

(C) Let the total height be $H$. The stone is dropped from rest,so $u = 0$.
Using the equation of motion $H = \frac{1}{2}gt^2$,for $t = 5\, s$,we get $H = \frac{1}{2}g(5)^2 = \frac{25}{2}g$.
Distance covered in the first $3\, s$ is $h_1 = \frac{1}{2}g(3)^2 = \frac{9}{2}g$.
Remaining distance $h_2 = H - h_1 = \frac{25}{2}g - \frac{9}{2}g = \frac{16}{2}g = 8g$.
When the stone is stopped after $3\, s$,its velocity at that instant is $v = gt = 3g$. However,it is then allowed to fall again from rest (as it was stopped),so for the remaining distance $h_2$,the initial velocity $u' = 0$.
Using $h_2 = \frac{1}{2}gt'^2$,where $t'$ is the time to cover the remaining distance:
$8g = \frac{1}{2}gt'^2$
$16 = t'^2$
$t' = 4\, s$.

(A) $1$. Height reached by the balloon in $t = 2\,s$:
$h_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 4.9 \times (2)^2 = 9.8\,m$.
$2$. Velocity of the balloon at $t = 2\,s$:
$v = a t = 4.9 \times 2 = 9.8\,m/s$.
$3$. When the ball is released,it has an initial upward velocity $u = 9.8\,m/s$. It moves under gravity until its velocity becomes zero at maximum height $h_2$ relative to the release point:
$v^2 = u^2 - 2 g h_2 \implies 0 = (9.8)^2 - 2 \times 9.8 \times h_2 \implies h_2 = \frac{9.8}{2} = 4.9\,m$.
$4$. The greatest height above the ground is the sum of the height at release and the additional height gained by the ball:
$H = h_1 + h_2 = 9.8 + 4.9 = 14.7\,m$.

(B) Let the total time taken to fall from height $h$ be $n$ seconds.
Using the equation of motion for distance covered from rest: $h = \frac{1}{2}gn^2$ ... $(i)$
Distance covered in the last second ($n^{th}$ second) is given by $S_n = \frac{g}{2}(2n - 1)$.
According to the problem,$S_n = \frac{9h}{25}$.
Substituting the value of $h$ from $(i)$ into this expression:
$\frac{9}{25} \left( \frac{1}{2}gn^2 \right) = \frac{g}{2}(2n - 1)$
$\frac{9n^2}{25} = 2n - 1$
$9n^2 = 50n - 25$
$9n^2 - 50n + 25 = 0$
Solving the quadratic equation: $9n^2 - 45n - 5n + 25 = 0$
$9n(n - 5) - 5(n - 5) = 0$
$(9n - 5)(n - 5) = 0$
Since $n$ must be greater than $1$ second for the last second to exist,we take $n = 5 \ s$.
Now,substitute $n = 5$ into equation $(i)$:
$h = \frac{1}{2} \times 9.8 \times (5)^2$
$h = 4.9 \times 25 = 122.5 \ m$.

(C) When the body is dropped from the ascending balloon,it acquires the same initial velocity as the balloon.
Initial velocity $u = +12\, m/s$ (upwards).
Displacement $s = -81\, m$ (downwards from the point of release to the ground).
Acceleration $a = g = -10\, m/s^2$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$-81 = 12t + \frac{1}{2}(-10)t^2$
$-81 = 12t - 5t^2$
$5t^2 - 12t - 81 = 0$
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$t = \frac{12 \pm \sqrt{(-12)^2 - 4(5)(-81)}}{2(5)}$
$t = \frac{12 \pm \sqrt{144 + 1620}}{10}$
$t = \frac{12 \pm \sqrt{1764}}{10}$
$t = \frac{12 \pm 42}{10}$
Since time cannot be negative,we take $t = \frac{12 + 42}{10} = \frac{54}{10} = 5.4\, s$.

(B) Let the time interval between consecutive drops be $\Delta t$.
When the first drop reaches the ground,the total time elapsed is $t = 2\Delta t$ (since the third drop is just leaving,meaning two intervals have passed).
Using the equation of motion $h = \frac{1}{2}gt^2$,for the first drop: $5 = \frac{1}{2} \times 10 \times t^2$,which gives $t^2 = 1$,so $t = 1\,s$.
Since $t = 2\Delta t$,the interval $\Delta t = 0.5\,s$.
At the instant the first drop hits the ground,the second drop has been falling for $\Delta t = 0.5\,s$.
The distance covered by the second drop from the tap is $y = \frac{1}{2}g(\Delta t)^2 = \frac{1}{2} \times 10 \times (0.5)^2 = 5 \times 0.25 = 1.25\,m$.
The height of the second drop above the ground is $H = 5 - 1.25 = 3.75\,m$.

(C) Given: Initial velocity $u = 4.9 \ m/s$ (upwards is taken as positive),time $t = 3 \ s$,acceleration due to gravity $g = 9.8 \ m/s^2$ (downwards is negative).
Using the equation of motion for displacement: $s = ut + \frac{1}{2}at^2$.
Here,the displacement $s$ is the height of the tower $h$ in the downward direction,so $s = -h$.
The acceleration $a = -g = -9.8 \ m/s^2$.
Substituting the values: $-h = (4.9)(3) + \frac{1}{2}(-9.8)(3)^2$.
$-h = 14.7 - 4.9 \times 9$.
$-h = 14.7 - 44.1$.
$-h = -29.4$.
Therefore,$h = 29.4 \ m$.

(A) The distance covered by a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{g}{2}(2n - 1)$.
Since the body starts from rest,the initial velocity $u = 0$.
Therefore,the distance covered in the $n^{th}$ second is $S_n = \frac{g}{2}(2n - 1)$.
For the first second $(n=1)$: $S_1 = \frac{g}{2}(2(1) - 1) = \frac{g}{2}(1) = \frac{g}{2}$.
For the second second $(n=2)$: $S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3) = \frac{3g}{2}$.
For the third second $(n=3)$: $S_3 = \frac{g}{2}(2(3) - 1) = \frac{g}{2}(5) = \frac{5g}{2}$.
Taking the ratio $S_1 : S_2 : S_3$,we get $\frac{g}{2} : \frac{3g}{2} : \frac{5g}{2} = 1 : 3 : 5$.

(B) When a stone is dropped from an ascending balloon,it inherits the velocity of the balloon at the moment of release. Let the height be $h$ and the acceleration due to gravity be $g$. The equation of motion for the stone is given by $h = -ut + \frac{1}{2}gt^2$,where $u$ is the upward velocity of the balloon.
Rearranging this gives a quadratic equation in time $t$: $\frac{1}{2}gt^2 - ut - h = 0$.
Solving for $t$ using the quadratic formula: $t = \frac{u + \sqrt{u^2 + 2gh}}{g}$.
Since $g$ and $h$ are constant for all stones,the time $t$ depends on the initial upward velocity $u$. As $u$ increases,the time $t$ taken to reach the ground increases.
Given velocities are $U_P = U$,$U_Q = 4U$,and $U_R = 8U$. Since $U_P < U_Q < U_R$,the stone with the smallest initial upward velocity $(U_P)$ will take the least amount of time to reach the ground.
Therefore,the stone from balloon $P$ reaches the ground first.

(B) At maximum height,the final velocity $v = 0$.
Using the first equation of motion,$v = u + at$:
$0 = u - gT \Rightarrow u = gT$.
To find the time $t$ when the velocity is $u/2$:
$v = u - gt$
$\frac{u}{2} = u - gt$
$gt = \frac{u}{2}$
Substituting $u = gT$ into the equation:
$gt = \frac{gT}{2} \Rightarrow t = \frac{T}{2}$.
Thus,the body acquires a velocity of $u/2$ at time $t = T/2$.

(C) For the first body,using the equation of motion $v^2 = u^2 + 2gh$:
Given $u = 0$,$v = 3 \ km/h$.
So,$3^2 = 0^2 + 2gh \implies 2gh = 9 \ (km/h)^2$.
For the second body,the initial speed $u = 4 \ km/h$ and it is dropped from the same height $h$.
Using the same equation $v'^2 = u^2 + 2gh$:
$v'^2 = (4)^2 + 2gh$.
Substituting $2gh = 9$:
$v'^2 = 16 + 9 = 25$.
$v' = \sqrt{25} = 5 \ km/h$.
Therefore,the final velocity is $5 \ km/h$.

(C) The distance covered in the $n^{th}$ second for an object projected vertically upwards is given by the formula: $h_{n^{th}} = u - \frac{g}{2}(2n - 1)$.
For the $5^{th}$ second $(n = 5)$: $h_{5^{th}} = u - \frac{10}{2}(2 \times 5 - 1) = u - 5(9) = u - 45$.
For the $6^{th}$ second $(n = 6)$: $h_{6^{th}} = u - \frac{10}{2}(2 \times 6 - 1) = u - 5(11) = u - 55$.
According to the problem,the distance covered in the $5^{th}$ second is twice the distance covered in the $6^{th}$ second:
$h_{5^{th}} = 2 \times h_{6^{th}}$.
Substituting the values:
$u - 45 = 2(u - 55)$.
$u - 45 = 2u - 110$.
$u = 110 - 45 = 65 \, m/s$.

(A) Let the first ball be $B_1$ and the second ball be $B_2$.
$B_1$ is dropped at $t = 0$. The distance covered by $B_1$ in $3$ seconds is $s_1 = \frac{1}{2} g (3)^2 = 4.5g$.
$B_2$ is dropped at $t = 1$ second. Therefore,at $t = 3$ seconds,$B_2$ has been falling for $2$ seconds. The distance covered by $B_2$ is $s_2 = \frac{1}{2} g (2)^2 = 2g$.
Taking $g = 10 \ m/s^2$,we get $s_1 = 45 \ m$ and $s_2 = 20 \ m$.
The distance between them is $s_1 - s_2 = 45 - 20 = 25 \ m$.

(B) Let the upward direction be positive and the downward direction be negative.
Initial velocity $u = +4.9 \, m/s$.
Time $t = 2 \, s$.
Acceleration due to gravity $g = -9.8 \, m/s^2$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$h = (4.9)(2) + \frac{1}{2}(-9.8)(2)^2$
$h = 9.8 - 4.9 \times 4$
$h = 9.8 - 19.6$
$h = -9.8 \, m$.
The negative sign indicates that the displacement is $9.8 \, m$ below the starting point.
Therefore,the height of the bridge is $9.8 \, m$.

(B) Using the principle of conservation of energy or kinematic equations,we can find the final velocity.
Let the upward direction be positive. The initial velocity $u = 20\, m/s$ and the displacement $s = -200\, m$ (since it lands below the starting point).
The acceleration due to gravity $g = -9.8\, m/s^2$.
Using the equation $v^2 = u^2 + 2as$:
$v^2 = (20)^2 + 2(-9.8)(-200)$
$v^2 = 400 + 3920 = 4320$
$v = \sqrt{4320} \approx 65.7\, m/s$.
Rounding to the nearest integer,the speed is approximately $65\, m/s$.

(B) Let the body start from rest at point $A$ with initial velocity $u = 0$.
For the motion from $A$ to $B$ (falling through height $h$):
Using the equation of motion $v^2 = u^2 + 2as$,where $u = 0$,$a = g$,and $s = h$:
$v^2 = 0 + 2gh$ ---$(i)$
For the motion from $A$ to $C$ (falling through total height $x$ to reach velocity $2v$):
Using the same equation $v_f^2 = u^2 + 2as$,where $v_f = 2v$,$u = 0$,$a = g$,and $s = x$:
$(2v)^2 = 0 + 2gx$
$4v^2 = 2gx$ ---(ii)
Dividing equation (ii) by equation $(i)$:
$\frac{4v^2}{v^2} = \frac{2gx}{2gh}$
$4 = \frac{x}{h}$
$x = 4h$
Therefore,the total distance from the starting point is $4h$.

(D) When a body is projected upwards with an initial velocity $u$,it undergoes constant deceleration due to gravity $(g)$.
At the maximum height,its velocity becomes $0$.
As it falls back down to the point of projection,it undergoes constant acceleration due to gravity $(g)$ over the same displacement.
Using the kinematic equation $v^2 = u^2 + 2as$,where $a = -g$ for the upward journey and $a = g$ for the downward journey,the final velocity $v$ at the starting point will have the same magnitude as the initial velocity $u$,but in the opposite direction.
Therefore,the speed of the body upon returning to the point of projection is $v = u$.

(D) The time of flight $T$ for a body projected vertically upwards is given by the formula $T = \frac{2u}{g}$.
Given that the total time taken to return to the starting point is $T = 4 \, s$ and the acceleration due to gravity is $g = 10 \, m/s^2$.
Substituting these values into the formula:
$4 = \frac{2u}{10}$
$40 = 2u$
$u = 20 \, m/s$.
Therefore,the initial velocity $u$ is $20 \, m/s$.

(B) For an object falling from rest,the distance $h$ covered in time $t$ is given by the equation of motion: $h = \frac{1}{2}gt^2$.
From this,we can express time $t$ as: $t = \sqrt{\frac{2h}{g}}$.
For heights $h_1$ and $h_2$ with corresponding times $t_1$ and $t_2$,we have:
$t_1 = \sqrt{\frac{2h_1}{g}}$ and $t_2 = \sqrt{\frac{2h_2}{g}}$.
Taking the ratio of $t_1$ to $t_2$:
$\frac{t_1}{t_2} = \frac{\sqrt{\frac{2h_1}{g}}}{\sqrt{\frac{2h_2}{g}}} = \sqrt{\frac{h_1}{h_2}} = \frac{\sqrt{h_1}}{\sqrt{h_2}}$.
Thus,the ratio is $\sqrt{h_1} : \sqrt{h_2}$.

(B) For a body thrown vertically upwards,the time taken to reach the maximum height (time of ascent) is equal to the time taken to fall back to the ground from that maximum height (time of descent),provided the air resistance is negligible.
Given that the time of ascent is $5\,sec$,the time of descent will also be $5\,sec$.
Therefore,the body will reach the ground from the maximum height position in $5\,sec$.

(B) Given,time of ascent $t = 6 \; s$. Since $t = u/g$,we have $u = 6 \times 10 = 60 \; m/s$ (taking $g = 10 \; m/s^2$).
Distance travelled in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
For the first second $(n = 1)$: $S_1 = 60 + \frac{-10}{2}(2(1) - 1) = 60 - 5 = 55 \; m$.
For the seventh second $(n = 7)$: $S_7 = 60 + \frac{-10}{2}(2(7) - 1) = 60 - 5(13) = 60 - 65 = -5 \; m$.
The magnitude of distance is $5 \; m$.
The ratio of the distances is $55:5 = 11:1$.

(B) Let the initial velocity be $u$ and the maximum height be $H$. The maximum height is given by $H = \frac{u^2}{2g}$.
At height $h = \frac{H}{2}$,the velocity $v = 10 \; m/s$.
Using the equation of motion $v^2 = u^2 - 2gh$:
$(10)^2 = u^2 - 2g \left( \frac{H}{2} \right)$
$100 = u^2 - gH$
Substitute $H = \frac{u^2}{2g}$ into the equation:
$100 = u^2 - g \left( \frac{u^2}{2g} \right)$
$100 = u^2 - \frac{u^2}{2} = \frac{u^2}{2}$
$u^2 = 200 \; m^2/s^2$.
Now,calculate the maximum height $H$:
$H = \frac{u^2}{2g} = \frac{200}{2 \times 10} = \frac{200}{20} = 10 \; m$.

(C) The maximum height $H$ attained by a body thrown vertically upwards with initial velocity $u$ is given by the formula $H = \frac{u^2}{2g}$.
From this formula,it is clear that the maximum height $H$ is independent of the mass of the body.
Given that the initial height $H_1 = 20\,m$ for initial velocity $u_1 = u$.
For the second body,the initial velocity $u_2 = 2u$.
The new maximum height $H_2$ is given by $H_2 = \frac{(2u)^2}{2g} = 4 \times \frac{u^2}{2g} = 4H_1$.
Substituting the value of $H_1$,we get $H_2 = 4 \times 20\,m = 80\,m$.

(A) $1$. Velocity of the balloon at $t = 8\, s$: $v = a \times t = 1.25 \times 8 = 10\, m/s$ (upward).
$2$. Height of the balloon at $t = 8\, s$: $s = \frac{1}{2} \times a \times t^2 = \frac{1}{2} \times 1.25 \times 8^2 = 40\, m$.
$3$. When the stone is released,its initial velocity is $u = 10\, m/s$ (upward) and its initial position is $h = 40\, m$ above the ground.
$4$. Using the equation of motion for displacement: $s = ut + \frac{1}{2}at^2$. Here,$s = -40\, m$ (downward),$u = 10\, m/s$,$a = -g = -10\, m/s^2$.
$-40 = 10t - 5t^2 \implies 5t^2 - 10t - 40 = 0 \implies t^2 - 2t - 8 = 0$.
Solving the quadratic equation: $(t - 4)(t + 2) = 0$. Since $t > 0$,$t = 4\, s$.
$5$. The displacement of the stone from the release point to the ground is $40\, m$ (downward).
$6$. The stone moves upward until its velocity becomes zero: $v^2 - u^2 = 2as \implies 0 - 10^2 = 2(-10)d \implies d = 5\, m$.
$7$. Total distance covered = $40\, m$ (upward) + $5\, m$ (upward) + $5\, m$ (downward) = $50\, m$.

(D) At the highest point of vertical motion,the body momentarily comes to rest,so the velocity $v = 0$.
However,the acceleration due to gravity $g$ continues to act downwards throughout the motion.
Using the kinematic equation $v^2 = u^2 - 2gH$,at the highest point $v = 0$,so $0 = u^2 - 2gH_{\max}$.
Solving for $H_{\max}$,we get $H_{\max} = u^2 / 2g$.
Therefore,option $D$ is the correct statement.

(D) Given: Maximum height $h = 20 \; m$,acceleration due to gravity $g = 10 \; m/s^2$.
At the highest point,the final velocity $v = 0$.
Using the equation of motion $v^2 = u^2 - 2gh$:
$0^2 = u^2 - 2 \times 10 \times 20$
$u^2 = 400$
$u = 20 \; m/s$.
Now,the total time of flight $T$ is given by $T = \frac{2u}{g}$:
$T = \frac{2 \times 20}{10} = 4 \; s$.
Therefore,the initial velocity is $20 \; m/s$ and the time in the air is $4 \; s$.

(D) For a particle moving under gravity,if it passes the same height $h$ at two different times $t_1$ and $t_2$,the height $h$ is given by the formula:
$h = \frac{1}{2} g t_1 t_2$
Given $t_1 = 2 \; s$ and $t_2 = 10 \; s$:
$h = \frac{1}{2} \times g \times 2 \times 10$
$h = 10g$
Thus,the correct option is $D$.

(B) The distance covered in the $n^{th}$ second is given by the formula: $S_n = u + \frac{g}{2}(2n - 1)$.
Given: Initial velocity $u = 10\; m/s$,acceleration $g = 10\; m/s^2$.
For the $3^{rd}$ second $(n=3)$:
$S_3 = 10 + \frac{10}{2}(2 \times 3 - 1) = 10 + 5(5) = 10 + 25 = 35\; m$.
For the $2^{nd}$ second $(n=2)$:
$S_2 = 10 + \frac{10}{2}(2 \times 2 - 1) = 10 + 5(3) = 10 + 15 = 25\; m$.
The ratio of the distances is $\frac{S_3}{S_2} = \frac{35}{25} = \frac{7}{5}$.

(C) Using the third equation of motion: $v^2 = u^2 + 2gh$,where $u$ is the initial speed,$g$ is the acceleration due to gravity,and $h$ is the height of the building.
For ball $A$ thrown upwards: The initial velocity is $u$ upwards. It will rise to a certain height and then fall back to the ground level. When it passes the point of projection on its way down,its speed will be $u$ downwards. Thus,it reaches the ground with a speed $v_A = \sqrt{u^2 + 2gh}$.
For ball $B$ thrown downwards: The initial velocity is $u$ downwards. It reaches the ground with a speed $v_B = \sqrt{u^2 + 2gh}$.
Since both $u$,$g$,and $h$ are the same for both balls,we have $v_A = v_B$.

(B) Let the first ball dropped from the top travel a distance $h_1$ in time $t$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$,we get:
$h_1 = \frac{1}{2}gt^2$
Let the second ball thrown from the bottom travel a distance $h_2$ in the same time $t$. Using $s = ut - \frac{1}{2}gt^2$,where $u = 50 \, m/s$,we get:
$h_2 = 50t - \frac{1}{2}gt^2$
Since the total height of the tower is $100 \, m$,the sum of the distances covered by both balls when they meet is equal to the height of the tower:
$h_1 + h_2 = 100$
Substituting the expressions for $h_1$ and $h_2$:
$\frac{1}{2}gt^2 + (50t - \frac{1}{2}gt^2) = 100$
$50t = 100$
$t = 2 \, s$
Thus,the balls will cross each other after $2 \, s$.

(B) Given: Initial velocity $u = 19.6 \ m/s$,acceleration due to gravity $g = 9.8 \ m/s^2$,and final velocity at maximum height $v = 0 \ m/s$.
Using the third equation of motion: $v^2 = u^2 - 2gH$.
Substituting the values: $0^2 = (19.6)^2 - 2 \times 9.8 \times H$.
$0 = 384.16 - 19.6 \times H$.
$19.6 \times H = 384.16$.
$H = \frac{384.16}{19.6} = 19.6 \ m$.
Therefore,the maximum height reached is $19.6 \ m$.

(B) Let the height of the minaret be $H$ and the total time taken by the body to fall from the top to the ground be $T$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g = 10 \ m/s^2$:
$H = \frac{1}{2}gT^2 = 5T^2$ ... $(i)$
In the last $2 \ s$,the body travels $40 \ m$. This means that in $(T - 2) \ s$,the body travels a distance of $(H - 40) \ m$.
$(H - 40) = \frac{1}{2}g(T - 2)^2 = 5(T - 2)^2$ ... (ii)
Substitute $H = 5T^2$ from $(i)$ into (ii):
$5T^2 - 40 = 5(T^2 - 4T + 4)$
$5T^2 - 40 = 5T^2 - 20T + 20$
$20T = 60$
$T = 3 \ s$
Now,substitute $T = 3 \ s$ into equation $(i)$:
$H = 5(3)^2 = 5 \times 9 = 45 \ m$.
Thus,the height of the minaret is $45 \ m$.