Motion Under Gravity Questions in English

(B) Given: Height of tower $h = 125 \ m$,initial velocity $u = 0$,acceleration $g = 10 \ m/s^2$.
First,calculate the total time $t$ taken to reach the ground using $h = ut + \frac{1}{2}gt^2$:
$125 = 0 + \frac{1}{2} \times 10 \times t^2$
$125 = 5t^2 \Rightarrow t^2 = 25 \Rightarrow t = 5 \ s$.
The distance covered in the last second is the distance covered in the total time minus the distance covered in $(t-1)$ seconds.
Distance covered in $(5-1) = 4 \ s$ is $h' = \frac{1}{2} \times 10 \times (4)^2 = 5 \times 16 = 80 \ m$.
Distance covered in the last second $= 125 - 80 = 45 \ m$.
According to the problem,this distance is $x \%$ of the total height:
$45 = \frac{x}{100} \times 125$
$x = \frac{45 \times 100}{125} = \frac{4500}{125} = 36$.

(B) Given: Initial velocity $u = 35 \ m/s$,acceleration $a = -g = -10 \ m/s^2$.
Using the first equation of motion $v = u + at$:
At $t_1 = 3 \ s$,the velocity is $v_1 = 35 - 10(3) = 35 - 30 = 5 \ m/s$.
The speed is $|v_1| = 5 \ m/s$.
At $t_2 = 4 \ s$,the velocity is $v_2 = 35 - 10(4) = 35 - 40 = -5 \ m/s$.
The speed is $|v_2| = |-5| = 5 \ m/s$.
Therefore,the ratio of the speeds at $3 \ s$ and $4 \ s$ is $\frac{|v_1|}{|v_2|} = \frac{5}{5} = 1: 1$.

(B) The distance travelled in the $n^{\text{th}}$ second is given by the formula:
$S_n = u + \frac{a}{2}(2n - 1)$
For a freely falling body,the initial velocity $u = 0$ and acceleration $a = g$.
Therefore,the distance travelled in the $n^{\text{th}}$ second is $S_n = \frac{g}{2}(2n - 1)$.
For the first second $(n = 1)$:
$S_1 = \frac{g}{2}(2(1) - 1) = \frac{g}{2}$
For the second second $(n = 2)$:
$S_2 = \frac{g}{2}(2(2) - 1) = \frac{3g}{2}$
For the third second $(n = 3)$:
$S_3 = \frac{g}{2}(2(3) - 1) = \frac{5g}{2}$
The ratio of displacements is $S_1 : S_2 : S_3 = \frac{g}{2} : \frac{3g}{2} : \frac{5g}{2} = 1 : 3 : 5$.

(B) When a body is projected vertically upwards, it experiences a constant downward acceleration due to gravity, denoted by $g$, throughout its entire flight.
At the highest point of its trajectory, the velocity of the body becomes $0 \,m/s$ momentarily.
However, the acceleration remains constant and acts downwards towards the center of the Earth.
Therefore, the acceleration at the highest point is equal to the acceleration due to gravity, which is approximately $9.8 \,m/s^2$ downwards.

(C) The equation of motion for height $H$ at time $t$ is given by $H = ut - \frac{1}{2}gt^2$.
Since the ball passes the same height $H$ at $t_1 = 2 \,s$ and $t_2 = 10 \,s$,we have:
$H = u(2) - \frac{1}{2}g(2)^2 = 2u - 2g$
$H = u(10) - \frac{1}{2}g(10)^2 = 10u - 50g$
Equating the two expressions for $H$:
$2u - 2g = 10u - 50g$
$8u = 48g$
$u = 6g = 6 \times 9.8 = 58.8 \,m/s$
Now,substitute $u$ back into the expression for $H$:
$H = 2(58.8) - 2(9.8) = 117.6 - 19.6 = 98 \,m$.

(B) Given that the height achieved in $1 \,s$ is $h=25 \,m$. Let $u$ be the initial velocity in the upward direction. Using the equation of motion,$h=ut-\frac{1}{2}gt^2$.
$25=u(1)-\frac{1}{2}(10)(1)^2$
$25=u-5 \Rightarrow u=30 \,m/s$.
The final velocity of the ball at maximum height becomes zero. The time taken to reach the maximum height is $v=u-gt \Rightarrow 0=30-10t \Rightarrow t=3 \,s$.
Now,the distance travelled in $2 \,s$ is the displacement in $2 \,s$ since the ball is still moving upwards:
$d_1=u(2)-\frac{1}{2}g(2)^2 = 30(2)-5(4) = 60-20 = 40 \,m$.
The maximum height achieved by the ball in $3 \,s$ is:
$H=u(3)-\frac{1}{2}g(3)^2 = 30(3)-5(9) = 90-45 = 45 \,m$.
The displacement after $4 \,s$ is:
$S_2=u(4)-\frac{1}{2}g(4)^2 = 30(4)-5(16) = 120-80 = 40 \,m$.
Since the ball has passed the maximum height at $t=3 \,s$,the total distance travelled in $4 \,s$ is $d_2=H+(H-S_2) = 45+(45-40) = 50 \,m$.
The ratio is $\frac{d_1}{d_2} = \frac{40}{50} = \frac{4}{5}$.

(D) At maximum height, the final velocity $v = 0$. Using the first equation of motion $v = u - gt$, we have $0 = u - 10 \times 5$, which gives $u = 50 \,m/s$.
Distance travelled in the $n^{th}$ second is given by $s_n = u + \frac{a}{2}(2n - 1)$.
For the $2^{nd}$ second $(n=2)$: $s_2 = 50 - \frac{10}{2}(2 \times 2 - 1) = 50 - 5(3) = 35 \,m$.
For the $7^{th}$ second $(n=7)$: $s_7 = 50 - \frac{10}{2}(2 \times 7 - 1) = 50 - 5(13) = 50 - 65 = -15 \,m$. The magnitude of distance is $15 \,m$.
The ratio of distance in the $2^{nd}$ second to the $7^{th}$ second is $\frac{35}{15} = \frac{7}{3}$.

(A) Let the initial position be $A$ and the point where the ball is caught be $B$. The highest point is $P$.
Initial speed $u = 20 \,m/s$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$ for the displacement $s = 5 \,m$:
$5 = 20t - \frac{1}{2}(10)t^2$
$5 = 20t - 5t^2$
Dividing by $5$:
$1 = 4t - t^2$
$t^2 - 4t + 1 = 0$
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$t = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \,s$.
Since the ball is caught on its way down, we take the larger time value:
$t = 2 + \sqrt{3} \,s$.

(B) For a freely falling body,the equation of motion for velocity is given by $v = u + gt$.
Since the initial speed $u = 0$,the equation becomes $v = gt$.
Here,$g$ is the acceleration due to gravity,which is a constant ($g \approx 9.8 \ m/s^2$ or $10 \ m/s^2$).
This equation $v = gt$ is of the form $y = mx$,where $y$ represents speed,$x$ represents time,and $m = g$ is the slope.
Since there is no constant term (intercept),the graph is a straight line passing through the origin $(0, 0)$.
Therefore,the correct option is $B$.

(A) According to the question, the ball reaches the top of a tower at time $t_1 = 3 \,s$ and further reaches its maximum height $2 \,s$ later.
Therefore, the total time taken by the ball to reach the maximum height is $t = t_1 + t_2 = 3 + 2 = 5 \,s$.
If $u$ is the initial velocity of the ball at $t = 0$, then from the first equation of motion $(v = u - gt)$:
$0 = u - 10 \times 5 \implies u = 50 \,m/s$.
If $h$ is the height of the tower, then from the second equation of motion $(h = ut - \frac{1}{2}gt^2)$:
$h = 50 \times 3 - \frac{1}{2} \times 10 \times 3^2 = 150 - 45 = 105 \,m$.
Hence, the height of the tower is $105 \,m$.

(A) At the maximum height, the final velocity of the object is $v = 0 \,m/s$.
Let the time taken to reach the maximum height be $t_{max}$.
We know that the velocity of an object at time $t$ before reaching the maximum height is equal to the velocity it gains in time $t$ while falling from the maximum height (due to symmetry).
Here, $t = 0.5 \,s$ and the acceleration due to gravity is $g = 9.8 \,m/s^2$.
Using the first equation of motion for the downward journey starting from rest at the maximum height:
$v = u + gt$
$v = 0 + (9.8 \,m/s^2) \times (0.5 \,s)$
$v = 4.9 \,m/s$.

(B) Let the distance $AB = BC = h$. The total distance $AC = 2h$.
Since the body falls freely from rest at $A$,the initial velocity $u = 0$.
Using the equation of motion $s = \frac{1}{2}gt^2$,the time $t_1$ taken to travel distance $AB = h$ is:
$t_1 = \sqrt{\frac{2h}{g}}$
The total time $T$ taken to travel distance $AC = 2h$ is:
$T = \sqrt{\frac{2(2h)}{g}} = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}}$
The time $t_2$ taken to travel distance $BC$ is the difference between the total time $T$ and the time $t_1$:
$t_2 = T - t_1 = 2\sqrt{\frac{h}{g}} - \sqrt{\frac{2h}{g}} = \sqrt{\frac{h}{g}}(2 - \sqrt{2})$
Now,the ratio $(t_2 / t_1)$ is:
$\frac{t_2}{t_1} = \frac{\sqrt{\frac{h}{g}}(2 - \sqrt{2})}{\sqrt{\frac{2h}{g}}} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1$

(B) Let the initial velocity be $u$. At maximum height $h$,the final velocity is $0$. Using $v = u - gt$,we get $0 = u - gt$,so $u = gt$.
The maximum height is $h = \frac{u^2}{2g} = \frac{(gt)^2}{2g} = \frac{gt^2}{2}$.
The body reaches the maximum height in time $t$.
While returning,the body falls from height $h$ to $h/2$. The distance covered is $h/2$.
Using $s = ut + \frac{1}{2}at^2$ for the downward motion starting from rest at the maximum height:
$\frac{h}{2} = 0 + \frac{1}{2}g(t')^2$,where $t'$ is the time taken to fall from the maximum height to half the maximum height.
Substituting $h = \frac{gt^2}{2}$,we get $\frac{gt^2}{4} = \frac{1}{2}g(t')^2$.
Solving for $t'$,we get $(t')^2 = \frac{t^2}{2}$,so $t' = \frac{t}{\sqrt{2}}$.
The total time from projection is $T = t + t' = t + \frac{t}{\sqrt{2}} = \left(1 + \frac{1}{\sqrt{2}}\right)t$.

(D) When a body reverses its direction,it comes to rest for a moment,for example,a body thrown upward at the topmost point. Hence,the reason statement is correct.
In the same example,at the topmost point,the velocity of the particle is $0$,but the acceleration of the particle is equal to the acceleration due to gravity $(g = 9.8 \ m/s^2)$,which is non-zero.
Therefore,the assertion statement is false.

(A) For a body falling freely under gravity,the initial velocity $u = 0$. The distance covered in time $t$ is given by $h = \frac{1}{2}gt^2$.
In the first $4$ seconds,the distance covered is $d_1 = \frac{1}{2}g(4)^2 = 8g$.
The total distance covered in $12$ seconds ($4$ seconds + $8$ seconds) is $h_{12} = \frac{1}{2}g(12)^2 = 72g$.
The distance covered in the next $8$ seconds is $d_2 = h_{12} - d_1 = 72g - 8g = 64g$.
Therefore,the ratio $\frac{d_2}{d_1} = \frac{64g}{8g} = 8$.

(B) Let the first ball be dropped from rest from height $h$ at $t=0$ and the second ball be dropped from the same height $h$ at $t=1 \,s$.
The distance covered by the first ball at time $t$ is $H_1 = \frac{1}{2} g t^2$.
The second ball is dropped at $t=1 \,s$. Let the time elapsed after the second ball is dropped be $t_1$. Then $t = 1 + t_1$.
The distance covered by the first ball at time $t$ is $s_1 = \frac{1}{2} g t^2 = \frac{1}{2} g (1 + t_1)^2$.
The distance covered by the second ball at time $t_1$ is $s_2 = \frac{1}{2} g t_1^2$.
The distance between the two balls is given by $s_1 - s_2 = 10 \,m$.
Substituting the expressions: $\frac{1}{2} g (1 + t_1)^2 - \frac{1}{2} g t_1^2 = 10$.
Using $g = 10 \,m/s^2$: $5(1 + 2t_1 + t_1^2) - 5t_1^2 = 10$.
$5 + 10t_1 + 5t_1^2 - 5t_1^2 = 10$.
$10t_1 = 5$.
$t_1 = 0.5 \,s$.
The total time $t = 1 + t_1 = 1 + 0.5 = 1.5 \,s$.

(A) Let the balls meet at a distance $h$ from the top after time $t$.
For ball-$1$ (downward motion):
$h = u_1 t + \frac{1}{2} g t^2 = 0 \cdot t + \frac{1}{2} (10) t^2 = 5 t^2$ ... $(i)$
For ball-$2$ (upward motion):
The distance covered by ball-$2$ is $(21 - h)$.
$(21 - h) = u_2 t - \frac{1}{2} g t^2 = 14 t - 5 t^2$ ... (ii)
Adding equations $(i)$ and (ii):
$h + (21 - h) = 5 t^2 + 14 t - 5 t^2$
$21 = 14 t$
$t = \frac{21}{14} = 1.5 \,s$
Substituting $t = 1.5 \,s$ into equation $(i)$:
$h = 5 \times (1.5)^2 = 5 \times 2.25 = 11.25 \,m = \frac{45}{4} \,m$
Thus,ball-$1$ will have dropped $\frac{45}{4} \,m$ when it passes ball-$2$.

(A) Let the two stones meet at a height $h$ from the ground after time $t$.
For the first stone dropped from $100 \ m$:
The height of the first stone at time $t$ is given by $y_1 = 100 - \frac{1}{2}gt^2$.
For the second stone projected upwards from the ground:
The height of the second stone at time $t$ is given by $y_2 = 25t - \frac{1}{2}gt^2$.
Since they meet at the same height,$y_1 = y_2$:
$100 - \frac{1}{2}gt^2 = 25t - \frac{1}{2}gt^2$
$100 = 25t$
$t = \frac{100}{25} = 4 \ s$.
Thus,the stones will be at the same height after $4 \ s$.

(D) Let the top of the window be the origin $(y=0)$. The stone is thrown upwards with $u = 8 \,m/s$.
At the maximum height $A$,the final velocity $v = 0$.
Using $v = u + at$:
$0 = 8 - 10t \implies t = 0.8 \,s$.
The height of point $A$ above the top of the window is $h = ut + \frac{1}{2}at^2 = 8(0.8) - \frac{1}{2}(10)(0.8)^2 = 6.4 - 3.2 = 3.2 \,m$.
Now,the stone falls from point $A$ to the bottom of the window. The total distance from $A$ to the bottom of the window is $H = 3.2 \,m + 1.8 \,m = 5.0 \,m$.
Time taken to fall from $A$ to the bottom of the window $(t_{total})$:
$H = \frac{1}{2}gt_{total}^2 \implies 5.0 = \frac{1}{2}(10)t_{total}^2 \implies t_{total}^2 = 1 \implies t_{total} = 1 \,s$.
The time taken to cross the window during the downward journey is the difference between the time to reach the bottom and the time to reach the top of the window from $A$:
$\Delta t = t_{total} - t = 1 \,s - 0.8 \,s = 0.2 \,s$.

(A) The time taken by the first ball to reach the ground is given by $t_1 = \sqrt{\frac{2h}{g}}$.
Substituting the values,$t_1 = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 \ s$.
The first ball hits the ground at $t = 10 \ s$.
The second ball is released $1 \ s$ later,so it has been in motion for $t_2 = 10 - 1 = 9 \ s$.
The distance traveled by the first ball is $s_1 = 500 \ m$ (as it hits the ground).
The distance traveled by the second ball in $9 \ s$ is $s_2 = \frac{1}{2} \times g \times t_2^2$.
$s_2 = \frac{1}{2} \times 10 \times 9^2 = 5 \times 81 = 405 \ m$.
The distance between the two balls is $s_1 - s_2 = 500 - 405 = 95 \ m$.

(B) Let the velocity of the particle at point $B$ be $v$.
Since the body is falling freely,the acceleration is $g$.
For the interval $BC$,the time taken is $t_1 = 2 \ s$. Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$BC = v(2) + \frac{1}{2}g(2)^2 = 2v + 2g$ --- $(i)$
For the interval $BD$,the total time is $t_2 = 2 + 1 = 3 \ s$.
$BD = BC + CD = 2BC$ (since $BC = CD$).
$BD = v(3) + \frac{1}{2}g(3)^2 = 3v + 4.5g$
Since $BD = 2BC$,we have:
$2(2v + 2g) = 3v + 4.5g$
$4v + 4g = 3v + 4.5g$
$v = 0.5g$
Now,for the interval $AB$,let the time taken be $t$. Since the body falls freely from $A$,its initial velocity $u = 0$.
$v = u + at \Rightarrow 0.5g = 0 + gt$
$t = 0.5 \ s$.

(B) Let the initial velocity of the body be $u$. At the maximum height $H$, the final velocity $v = 0$. The formula for maximum height is $H = \frac{u^2}{2g}$.
At a height $h$, the velocity is $v_h = 8 \,m/s$. Using the equation of motion $v^2 = u^2 - 2gh$, we have $8^2 = u^2 - 2gh$, which implies $u^2 = 64 + 2gh$.
However, the problem implies the body reaches a height $h$ where the speed is $8 \,m/s$. Assuming the body is projected from the ground, the maximum height $H$ is reached when the velocity becomes $0$. If we consider the motion from the point of height $h$ to the maximum height $H$, the initial velocity is $8 \,m/s$ and final velocity is $0$. The distance covered is $(H - h)$.
Using $v^2 = u^2 - 2as$, we get $0^2 = 8^2 - 2g(H - h)$, so $64 = 20(H - h)$, which gives $(H - h) = 3.2 \,m$.
If the question implies the body is at height $h$ and the speed is $8 \,m/s$, the additional height gained is $3.2 \,m$. Assuming the body was projected such that $h$ is the point where speed is $8 \,m/s$, the maximum height relative to that point is $3.2 \,m$.

(C) Let the time taken by the body to fall from the maximum height point $C$ to point $B$ be $t^{\prime}$.
Due to symmetry of motion,the time taken to go from $B$ to $C$ is equal to the time taken to fall from $C$ to $B$,which is $t^{\prime}$.
Given that the body is at height $H$ at times $t_1$ and $t_2$,we have $t_2 = t_1 + 2t^{\prime}$.
Thus,$t^{\prime} = \frac{t_2-t_1}{2}$.
The total time $T$ taken to reach the maximum height $C$ is $T = t_1 + t^{\prime} = t_1 + \frac{t_2-t_1}{2} = \frac{t_1+t_2}{2}$.
The maximum height $H_{\max}$ attained is given by the formula $H_{\max} = \frac{1}{2}gT^2$.
Substituting the value of $T$,we get $H_{\max} = \frac{1}{2}g\left(\frac{t_1+t_2}{2}\right)^2 = \frac{1}{2}g \cdot \frac{(t_1+t_2)^2}{4} = \frac{g(t_1+t_2)^2}{8}$.

(C) Let the body be projected with initial velocity $u$. The equation of motion for height $h$ is $h = ut - \frac{1}{2}gt^2$,which is a quadratic equation in $t$: $\frac{1}{2}gt^2 - ut + h = 0$.
Since $t_1$ and $t_2$ are the roots of this equation,the sum of roots is $t_1 + t_2 = \frac{u}{g/2} = \frac{2u}{g}$.
Thus,$u = \frac{g(t_1+t_2)}{2}$.
The maximum height $H$ reached by the body is given by $H = \frac{u^2}{2g}$.
Substituting the value of $u$: $H = \frac{1}{2g} \left[ \frac{g(t_1+t_2)}{2} \right]^2 = \frac{1}{2g} \cdot \frac{g^2(t_1+t_2)^2}{4} = \frac{g(t_1+t_2)^2}{8}$.
Alternatively,$H = 2g \left( \frac{t_1+t_2}{4} \right)^2 = 2g \cdot \frac{(t_1+t_2)^2}{16} = \frac{g(t_1+t_2)^2}{8}$.
Therefore,the correct option is $C$.

(C) Given: Height of the bridge $h = 45 \ m$. Initial velocity of the ball $u = 0 \ m \ s^{-1}$. Acceleration due to gravity $g = 10 \ m \ s^{-2}$.
Using the kinematic equation for vertical motion: $h = ut + \frac{1}{2}gt^2$.
Since $u = 0$,we have $45 = 0 + \frac{1}{2} \times 10 \times t^2$.
$45 = 5t^2 \implies t^2 = 9 \implies t = 3 \ s$.
The ball takes $3 \ s$ to reach the water surface.
During this time,the boat must travel the distance of $12 \ m$ to be at the point of impact.
Speed of the boat $v = \frac{\text{distance}}{\text{time}} = \frac{12 \ m}{3 \ s} = 4 \ m \ s^{-1}$.

(B) Let $t_A = 2 \ s$ be the time interval for boy $A$ and $t_B = 1 \ s$ be the time interval for boy $B$.
For a ball projected vertically,the time taken to reach the maximum height from a point at height $h$ is $t/2$.
Using $v = u + at$,at the highest point $v = 0$,so $u = g(t/2)$.
The height $h$ of the point from the ground is given by $h = u(t/2) - 1/2 g(t/2)^2 = g(t/2)^2 - 1/2 g(t/2)^2 = 1/2 g(t/2)^2 = 1/8 g t^2$.
For boy $A$: $h_A = 1/8 \times 10 \times (2)^2 = 5 \ m$.
For boy $B$: $h_B = 1/8 \times 10 \times (1)^2 = 1.25 \ m$.
The difference in vertical positions is $h_A - h_B = 5 \ m - 1.25 \ m = 3.75 \ m$.

(A) The work done by the force of gravity is given by $W = F \cdot S_n = mg \cdot S_n$,where $S_n$ is the distance traveled in the $n$th second.
Since $m$,$g$,and $F$ are constant,$W \propto S_n$.
The distance traveled in the $n$th second for an object starting from rest is given by $S_n = u + \frac{a}{2}(2n - 1)$.
Here,$u = 0$ and $a = g$,so $S_n = \frac{g}{2}(2n - 1)$.
For $n = 1, 2, 3$:
$S_1 = \frac{g}{2}(2(1) - 1) = \frac{g}{2}(1)$
$S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3)$
$S_3 = \frac{g}{2}(2(3) - 1) = \frac{g}{2}(5)$
The ratio of distances is $S_1 : S_2 : S_3 = 1 : 3 : 5$.
Therefore,the ratio of work done is $1 : 3 : 5$.

(C) Let the time interval between two consecutive drops be $t$.
Since the first drop strikes the floor when the sixth drop begins to fall,the total time taken by the first drop to reach the ground is $5t$.
Given the height $h = 5 \ m$ and $g = 10 \ m/s^2$,the time taken is $t_{total} = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 \ s$.
Thus,$5t = 1 \ s$,which implies $t = 0.2 \ s$.
At the instant the first drop strikes the ground,the fourth drop has been falling for a time $t_4 = (5 - 4)t = 1 \times 0.2 = 0.2 \ s$ is incorrect; let's re-evaluate:
The first drop is at $t=0$ (start),$t=1$ (ground).
The second drop is at $t=0.2$,the third at $t=0.4$,the fourth at $t=0.6$,the fifth at $t=0.8$,and the sixth at $t=1.0$.
At $t=1.0 \ s$,the fourth drop has been falling for $(1.0 - 0.6) = 0.4 \ s$.
Distance fallen by the fourth drop $h_4 = \frac{1}{2}gt_4^2 = \frac{1}{2} \times 10 \times (0.4)^2 = 5 \times 0.16 = 0.8 \ m$.
Height from the ground $= 5 - 0.8 = 4.2 \ m$.

(B) Let the points of motion be $A$ (jump),$B$ (parachute opens),$C$ (at $10 \ m$ height),and $D$ (ground).
$1.$ Motion from $A$ to $B$ (Free fall):
Initial velocity $u_A = 0$,time $t = 2 \ s$,acceleration $a = g = 10 \ m/s^2$.
Distance covered $x_1 = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times (2)^2 = 20 \ m$.
Velocity at $B$,$v_B = u_A + gt = 0 + 10 \times 2 = 20 \ m/s$.
$2.$ Motion from $B$ to $C$ (Deceleration):
Initial velocity $u_B = 20 \ m/s$,final velocity $v_C = 5 \ m/s$,acceleration $a = -3 \ m/s^2$.
Using $v^2 - u^2 = 2as$:
$(5)^2 - (20)^2 = 2(-3) x_2$
$25 - 400 = -6 x_2$
$-375 = -6 x_2$
$x_2 = \frac{375}{6} = 62.5 \ m$.
$3.$ Motion from $C$ to $D$:
Distance $x_3 = 10 \ m$.
Total height $H = x_1 + x_2 + x_3 = 20 + 62.5 + 10 = 92.5 \ m$.

(A) Let the ball fall a distance $h'$ from the top.
The velocity $v$ of the ball after falling a distance $h'$ is given by $v = \sqrt{2gh'}$.
The magnitude of acceleration due to gravity is $a = g = 10 \text{ m/s}^2$.
According to the problem,the magnitude of velocity equals the magnitude of acceleration: $|v| = |a|$.
Substituting the values,we get $\sqrt{2gh'} = g$.
Squaring both sides,we get $2gh' = g^2$.
Dividing by $g$,we get $2h' = g$.
Given $g = 10 \text{ m/s}^2$,we have $2h' = 10$,which gives $h' = 5 \text{ m}$.
The height above the ground $H$ is the total height minus the distance fallen: $H = 18 \text{ m} - 5 \text{ m} = 13 \text{ m}$.

(D) The distance travelled by a freely falling object is given by the equation $s = \frac{1}{2}gt^2$.
Since $g$ is constant,the distance $s$ is directly proportional to the square of the reaction time $t$ $(s \propto t^2)$.
Comparing the reaction times: $t_B (0.22 \text{ s}) > t_E (0.21 \text{ s}) > t_A (0.20 \text{ s}) > t_D (0.19 \text{ s}) > t_C (0.18 \text{ s})$.
Since $s \propto t^2$,the order of distances will be the same as the order of the squares of the reaction times.
Therefore,the distance travelled will follow the order: $B > E > A > D > C$.
Thus,option $D$ is correct.