Equations of Motion Questions in English

(D) Given:
Initial velocity,$u = 0\, m s^{-1}$
Acceleration,$a = 0.1\, m s^{-2}$
Time,$t = 2\, min = 120\, s$
$(a)$ To find the speed acquired $(v)$:
Using the first equation of motion,$v = u + at$
$v = 0 + (0.1\, m s^{-2} \times 120\, s)$
$v = 12\, m s^{-1}$
$(b)$ To find the distance travelled $(s)$:
Using the second equation of motion,$s = ut + \frac{1}{2}at^2$
$s = (0 \times 120) + \frac{1}{2} \times 0.1 \times (120)^2$
$s = 0 + 0.05 \times 14400$
$s = 720\, m$
Thus,the speed acquired is $12\, m s^{-1}$ and the distance travelled is $720\, m$.

(A) Initial speed of the train,$u = 90\, km/h = 90 \times \frac{5}{18} = 25\, m/s$.
Final speed of the train,$v = 0\, m/s$ (as the train comes to rest).
Acceleration,$a = -0.5\, m/s^2$.
Using the third equation of motion,$v^2 = u^2 + 2as$.
Substituting the values: $(0)^2 = (25)^2 + 2(-0.5)s$.
$0 = 625 - 1s$.
$s = 625\, m$.
The train will cover a distance of $625\, m$ before it comes to rest.

(A) Initial velocity of the trolley,$u = 0 \,m \,s^{-1} = 0 \,cm \,s^{-1}$.
Acceleration,$a = 2 \,m \,s^{-2} = 200 \,cm \,s^{-2}$.
Time,$t = 3 \,s$.
Using the first equation of motion,$v = u + at$.
Substituting the values: $v = 0 + (200 \,cm \,s^{-2} \times 3 \,s)$.
$v = 600 \,cm \,s^{-1}$.
Therefore,the velocity of the trolley after $3 \,s$ is $600 \,cm \,s^{-1}$.

(C) Initial velocity of the car,$u = 0 \,m s^{-1}$.
Acceleration,$a = 4 \,m s^{-2}$.
Time,$t = 10 \,s$.
We use the second equation of motion: $s = ut + \frac{1}{2}at^2$.
Substituting the values: $s = (0 \times 10) + \frac{1}{2} \times 4 \times (10)^2$.
$s = 0 + \frac{1}{2} \times 4 \times 100$.
$s = 2 \times 100 = 200 \,m$.
Therefore,the distance covered by the car in $10 \,s$ is $200 \,m$.

(D) Given: Initial velocity $u = 5 \, m s^{-1}$.
Final velocity at maximum height $v = 0 \, m s^{-1}$.
Acceleration $a = -10 \, m s^{-2}$ (since it acts downwards).
Using the equation of motion $v^2 - u^2 = 2as$:
$0^2 - (5)^2 = 2 \times (-10) \times s$
$-25 = -20s$
$s = \frac{-25}{-20} = 1.25 \, m$.
Using the equation of motion $v = u + at$:
$0 = 5 + (-10)t$
$10t = 5$
$t = 0.5 \, s$.
Thus,the height attained is $1.25 \, m$ and the time taken is $0.5 \, s$.

(D) Given: Initial velocity of the motorboat,$u = 0 \, m s^{-1}$.
Acceleration of the motorboat,$a = 3.0 \, m s^{-2}$.
Time interval,$t = 8.0 \, s$.
We use the second equation of motion: $s = ut + \frac{1}{2}at^2$.
Substituting the given values:
$s = (0 \times 8.0) + \frac{1}{2} \times 3.0 \times (8.0)^2$.
$s = 0 + \frac{1}{2} \times 3.0 \times 64$.
$s = 3.0 \times 32$.
$s = 96 \, m$.
Therefore,the boat travels $96 \, m$ during this time.

(C) Let the final velocity with which the ball strikes the ground be $v$ and the time taken be $t$.
Initial velocity of the ball,$u = 0 \, m s^{-1}$.
Distance or height of fall,$s = 20 \, m$.
Downward acceleration,$a = 10 \, m s^{-2}$.
Using the third equation of motion: $v^2 - u^2 = 2as$.
$v^2 = 2as + u^2$.
$v^2 = 2 \times 10 \times 20 + 0^2 = 400$.
Therefore,final velocity $v = \sqrt{400} = 20 \, m s^{-1}$.
Using the first equation of motion: $v = u + at$.
$20 = 0 + 10 \times t$.
Therefore,time $t = 20 / 10 = 2 \, s$.

(C) Given: Mass of the dumbbell,$m = 10\, kg$.
Distance covered,$s = 80\, cm = 0.8\, m$.
Acceleration,$a = 10\, m \,s^{-2}$.
Initial velocity,$u = 0\, m \,s^{-1}$.
Using the third equation of motion,$v^2 = u^2 + 2as$:
$v^2 = 0^2 + 2 \times 10 \times 0.8$
$v^2 = 16$
$v = 4\, m \,s^{-1}$.
Momentum $(p)$ is given by the product of mass and velocity: $p = m \times v$.
$p = 10\, kg \times 4\, m \,s^{-1} = 40\, kg\, m \,s^{-1}$.
Therefore,the dumbbell will transfer $40\, kg\, m \,s^{-1}$ of momentum to the floor.

(A) Time of ascent is equal to the time of descent. The ball takes a total of $6\, s$ for its upward and downward journey.
Hence,it has taken $3\, s$ to attain the maximum height.
Final velocity of the ball at the maximum height,$v = 0\, m/s$.
Acceleration due to gravity,$g = -9.8\, m/s^2$.
Using the equation of motion,$v = u + at$,we have:
$0 = u + (-9.8 \times 3)$
$u = 9.8 \times 3 = 29.4\, m/s$.
Hence,the ball was thrown upwards with a velocity of $29.4\, m/s$.
$(b)$ Let the maximum height attained by the ball be $h$.
Initial velocity during the upward journey,$u = 29.4\, m/s$.
Final velocity,$v = 0\, m/s$.
Acceleration due to gravity,$g = -9.8\, m/s^2$.
Using the equation of motion,$s = ut + \frac{1}{2}at^2$:
$h = 29.4 \times 3 - \frac{1}{2} \times 9.8 \times 3^2 = 88.2 - 44.1 = 44.1\, m$.
Hence,the maximum height is $44.1\, m$.
$(c)$ The ball attains the maximum height after $3\, s$. After attaining this height,it starts falling downwards.
In this case,initial velocity,$u = 0\, m/s$.
The position of the ball after $4\, s$ of the throw is given by the distance traveled by it during its downward journey in $4\, s - 3\, s = 1\, s$.
Using the equation of motion,$s = ut + \frac{1}{2}at^2$:
$s = 0 \times 1 + \frac{1}{2} \times 9.8 \times 1^2 = 4.9\, m$.
Now,the total height is $44.1\, m$.
This means the ball is $39.2\, m$ $(44.1\, m - 4.9\, m)$ above the ground after $4\, s$.

(B) When a body is thrown vertically upward with an initial velocity $u$,it reaches its maximum height $h$ where its final velocity $v$ becomes $0$.
According to the third equation of motion: $v^2 = u^2 + 2as$.
Here,$v = 0$,$a = -g$ (acceleration due to gravity acting downwards),and $s = h$.
Substituting these values: $0^2 = u^2 + 2(-g)h$.
$0 = u^2 - 2gh$.
$2gh = u^2$.
Therefore,the maximum height reached is $h = u^2 / (2g)$.

(N/A) When an object moves with a uniform velocity,the acceleration $(a)$ is $0$ and the final velocity $(v)$ is equal to the initial velocity $(u)$,i.e.,$v = u$.
$1$. The first equation of motion,$v = u + at$,becomes $v = u + (0)t$,which simplifies to $v = u$.
$2$. The second equation of motion,$s = ut + (1/2)at^2$,becomes $s = ut + (1/2)(0)t^2$,which simplifies to $s = ut$.
$3$. The third equation of motion,$v^2 - u^2 = 2as$,becomes $v^2 - u^2 = 2(0)s$,which simplifies to $v^2 - u^2 = 0$ or $v^2 = u^2$.

(D) Given that the object starts from rest,initial velocity $u = 0 \, m/s$.
For the first $2 \, s$,distance $s_1 = 20 \, m$ and time $t_1 = 2 \, s$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$20 = 0(2) + \frac{1}{2} a(2)^2 \Rightarrow 20 = 2a \Rightarrow a = 10 \, m/s^2$.
Since the acceleration is constant,we can calculate the velocity at any time $t$ using $v = u + at$.
For $t = 7 \, s$:
$v = 0 + (10 \times 7) = 70 \, m/s$.

(B) Given initial velocity,$u = 5 \times 10^{4} \, ms^{-1}$.
Acceleration,$a = 10^{4} \, ms^{-2}$.
$(i)$ Final velocity $v = 2u = 2 \times 5 \times 10^{4} = 10 \times 10^{4} \, ms^{-1}$.
Using the first equation of motion,$v = u + at$,we get $t = \frac{v - u}{a}$.
$t = \frac{10 \times 10^{4} - 5 \times 10^{4}}{10^{4}} = \frac{5 \times 10^{4}}{10^{4}} = 5 \, s$.
$(ii)$ Using the second equation of motion,$s = ut + \frac{1}{2}at^{2}$.
$s = (5 \times 10^{4}) \times 5 + \frac{1}{2} \times (10^{4}) \times (5)^{2}$.
$s = 25 \times 10^{4} + 12.5 \times 10^{4} = 37.5 \times 10^{4} \, m$.

(C) The distance travelled by an object in time $t$ with uniform acceleration $a$ and initial velocity $u$ is given by the equation of motion: $s = ut + \frac{1}{2}at^2$.
First,calculate the total distance travelled in $5 \, s$:
$s_5 = u(5) + \frac{1}{2}a(5)^2 = 5u + \frac{25}{2}a$ ........... $(i)$
Next,calculate the total distance travelled in $4 \, s$:
$s_4 = u(4) + \frac{1}{2}a(4)^2 = 4u + \frac{16}{2}a = 4u + 8a$ ........... $(ii)$
The distance travelled in the interval between the $4^{th}$ and $5^{th}$ second is the difference between the distance at $5 \, s$ and $4 \, s$:
Distance $= s_5 - s_4 = (5u + \frac{25}{2}a) - (4u + 8a)$
Distance $= (5u - 4u) + (12.5a - 8a) = u + 4.5a$
Distance $= (u + \frac{9}{2}a) \, m$.

For an object moving vertically upwards,the third equation of motion is given by $v^{2} = u^{2} + 2as$.
At the maximum height,the final velocity $v$ becomes $0$.
Substituting $v = 0$,$a = -g$,and $s = h$ into the equation: $0 = u^{2} - 2gh$.
Rearranging for height,we get $h = \frac{u^{2}}{2g}$.
For the first stone with initial velocity $u_{1}$,the maximum height reached is $h_{1} = \frac{u_{1}^{2}}{2g}$.
For the second stone with initial velocity $u_{2}$,the maximum height reached is $h_{2} = \frac{u_{2}^{2}}{2g}$.
Taking the ratio of the two heights: $\frac{h_{1}}{h_{2}} = \frac{u_{1}^{2} / 2g}{u_{2}^{2} / 2g} = \frac{u_{1}^{2}}{u_{2}^{2}}$.
Therefore,the ratio of the heights reached is $h_{1}: h_{2} = u_{1}^{2}: u_{2}^{2}$.

(N/A) The three equations of motion for a body moving with uniform acceleration are:
$(i)$ $v = u + at$
$(ii)$ $S = ut + \frac{1}{2}at^2$
$(iii)$ $v^2 = u^2 + 2aS$
Where:
$u$ = initial velocity
$v$ = final velocity
$a$ = uniform acceleration
$t$ = time taken
$S$ = displacement covered
Any two of these equations can be provided as the answer.

(A) Given: Time $t = 3 \ s$,Final velocity $v = 0 \ m s^{-1}$ (at maximum height),Acceleration $a = g = -9.8 \ m s^{-2}$ (acting downwards).
To find initial velocity $u$:
Using the first equation of motion: $v = u + at$
$0 = u + (-9.8) \times 3$
$u = 29.4 \ m s^{-1}$.
To find maximum height $h$:
Using the second equation of motion: $h = ut + \frac{1}{2}at^2$
$h = (29.4 \times 3) + \frac{1}{2} \times (-9.8) \times (3)^2$
$h = 88.2 - 44.1$
$h = 44.1 \ m$.

(N/A) Consider the velocity-time graph shown in the figure.
The distance $S$ covered by an object moving with uniform acceleration $a$ is equal to the area under the velocity-time graph,which is the area of the trapezium $OABD$.
Area of trapezium $OABD = \frac{(OA + BD) \times OD}{2}$
From the graph,we have:
$OA = u$ (initial velocity)
$BD = v$ (final velocity)
$OD = t$ (time taken)
Substituting these values,we get:
$S = \frac{(u + v) t}{2}$ --- (Equation $1$)
We know that for uniform acceleration,$v = u + at$,which can be rearranged to find time $t$:
$at = v - u$
$t = \frac{v - u}{a}$ --- (Equation $2$)
Substituting the value of $t$ from Equation $2$ into Equation $1$:
$S = \frac{(u + v)(v - u)}{2a}$
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$,we get:
$S = \frac{v^2 - u^2}{2a}$
Rearranging the terms,we get the final equation:
$v^2 - u^2 = 2aS$

(N/A) The velocity-time graph for a body moving with constant acceleration $a$ and initial velocity $u$ is shown in the figure.
The distance covered $S$ by the object in time $t$ is equal to the area under the velocity-time graph.
The area under the graph is the sum of the area of the rectangle $OACD$ and the area of the triangle $ABC$ on top of it.
The area of the rectangle $OACD$ is given by: $\text{Area} = \text{length} \times \text{breadth} = t \times u = ut$.
The area of the triangle $ABC$ is given by: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times (v - u)$.
Since the total distance $S$ is the sum of these areas,we have:
$S = ut + \frac{1}{2} \times t \times (v - u)$ $....(1)$
From the first equation of motion,we know that $v = u + at$,which implies $v - u = at$.
Substituting $(v - u) = at$ into equation $(1)$,we get:
$S = ut + \frac{1}{2} \times t \times (at)$
$S = ut + \frac{1}{2}at^{2}$

(N/A) Suppose the initial velocity of a body is $u$ and it is moving with uniform acceleration $a$ for time $t$. Let the final velocity be $v$ and the distance covered be $S$.
$(i)$ Acceleration is defined as the rate of change of velocity:
$a = \frac{v - u}{t}$
$at = v - u$
$v = u + at$
$(ii)$ The average velocity for uniform acceleration is given by:
$\bar{v} = \frac{u + v}{2} \quad \dots(1)$
Also, average velocity is total displacement divided by time:
$\bar{v} = \frac{S}{t} \quad \dots(2)$
Equating $(1)$ and $(2)$:
$\frac{u + v}{2} = \frac{S}{t} \implies S = \left( \frac{u + v}{2} \right) t \quad \dots(3)$
Substituting $v = u + at$ into equation $(3)$:
$S = \left( \frac{u + (u + at)}{2} \right) t$
$S = \left( \frac{2u + at}{2} \right) t$
$S = ut + \frac{1}{2}at^2$
$(iii)$ From equation $(3)$, $S = \left( \frac{u + v}{2} \right) t$. From $v = u + at$, we get $t = \frac{v - u}{a}$.
Substituting $t$ in equation $(3)$:
$S = \left( \frac{u + v}{2} \right) \left( \frac{v - u}{a} \right)$
$S = \frac{v^2 - u^2}{2a}$
$v^2 - u^2 = 2aS$

(A) The area under the velocity-time graph represents the distance travelled by the object. As shown in the figure,the area is the sum of the area of the rectangle $OACD$ and the area of the triangle $ABC$.
The area of the rectangle $OACD = \text{length} \times \text{breadth} = OC \times OA = t \times u = ut$.
The area of the triangle $ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times t \times (v - u)$.
Since $v = u + at$,we have $(v - u) = at$. Substituting this into the area of the triangle,we get:
Area of triangle $ABC = \frac{1}{2} \times t \times (at) = \frac{1}{2}at^2$.
Therefore,the total distance $S = \text{Area of rectangle } OACD + \text{Area of triangle } ABC = ut + \frac{1}{2}at^2$.
$(b)$ Given:
Initial velocity $u = 18 \text{ km h}^{-1} = 18 \times \frac{5}{18} \text{ m s}^{-1} = 5 \text{ m s}^{-1}$.
Final velocity $v = 36 \text{ km h}^{-1} = 36 \times \frac{5}{18} \text{ m s}^{-1} = 10 \text{ m s}^{-1}$.
Time $t = 5 \text{ s}$.
Acceleration $a = \frac{v - u}{t} = \frac{10 - 5}{5} = \frac{5}{5} = 1 \text{ m s}^{-2}$.
Distance $S = ut + \frac{1}{2}at^2 = (5 \times 5) + \frac{1}{2} \times 1 \times (5)^2 = 25 + 12.5 = 37.5 \text{ m}$.

(B) Given: Initial velocity $u = 0\, m s^{-1}$,acceleration $a = 4\, m s^{-2}$,and time $t = 20\, s$.
To find the distance $S$ covered,we use the second equation of motion:
$S = ut + \frac{1}{2}at^2$
Substituting the given values:
$S = (0 \times 20) + \frac{1}{2} \times 4 \times (20)^2$
$S = 0 + 2 \times 400$
$S = 800\, m$
Therefore,the car will cover a distance of $800\, m$ in $20\, s$.

(A) Given:
Initial velocity $u = 90\, km\, h^{-1} = 90 \times \frac{5}{18} = 25\, m\, s^{-1}$.
Final velocity $v = 0\, m\, s^{-1}$ (since the train comes to rest).
Acceleration $a = -0.5\, m\, s^{-2}$ (negative sign indicates retardation).
We need to find the distance $S$.
Using the third equation of motion: $v^{2} - u^{2} = 2aS$.
Substituting the values: $0^{2} - (25)^{2} = 2 \times (-0.5) \times S$.
$-625 = -1 \times S$.
$S = 625\, m$.
Therefore,the train will travel $625\, m$ before coming to rest.

(A) Given: Initial velocity $u = 40 \, m s^{-1}$,Final velocity $v = 0 \, m s^{-1}$,Acceleration $a = -2 \, m s^{-2}$.
Using the third equation of motion: $v^{2} = u^{2} + 2aS$.
Substituting the values: $0^{2} = (40)^{2} + 2(-2)S$.
$0 = 1600 - 4S$.
$4S = 1600$.
$S = 400 \, m$.
Since the calculated stopping distance is $400 \, m$ and the platform length is $400 \, m$,the train will stop exactly at the end of the platform.

(B) Given: Initial velocity $u = 0 \, m/s$,acceleration $a = 2.5 \, m s^{-2}$,and time $t = 4 \, s$.
To find the distance traveled $(S)$,we use the second equation of motion:
$S = ut + \frac{1}{2}at^2$
Substituting the given values into the equation:
$S = (0 \times 4) + \frac{1}{2} \times 2.5 \times (4)^2$
$S = 0 + 0.5 \times 2.5 \times 16$
$S = 1.25 \times 16 = 20 \, m$
Therefore,the girl travels $20 \, m$ in this time.

(N/A) Given: Initial velocity $u = 5\, m s^{-1}$,Final velocity $v = 10\, m s^{-1}$,Distance $S = 50\, m$.
$(i)$ To calculate acceleration $(a)$,we use the third equation of motion: $v^{2} - u^{2} = 2aS$.
Substituting the values: $(10)^{2} - (5)^{2} = 2 \times a \times 50$.
$100 - 25 = 100a$.
$75 = 100a$.
$a = \frac{75}{100} = 0.75\, m s^{-2}$.
$(ii)$ To calculate time $(t)$,we use the first equation of motion: $v = u + at$.
Substituting the values: $10 = 5 + 0.75 \times t$.
$10 - 5 = 0.75t$.
$5 = 0.75t$.
$t = \frac{5}{0.75} = 6.67\, s$.

(A) Given: Initial velocity $u = 90 \text{ km h}^{-1} = 90 \times \frac{5}{18} \text{ m s}^{-1} = 25 \text{ m s}^{-1}$.
Final velocity $v = 0 \text{ m s}^{-1}$ (as the truck comes to rest).
Distance $S = 25 \text{ m}$.
$(i)$ To find retardation $(a)$: Using the third equation of motion,$v^{2} - u^{2} = 2aS$.
$(0)^{2} - (25)^{2} = 2 \times a \times 25$.
$-625 = 50a$.
$a = -\frac{625}{50} = -12.5 \text{ m s}^{-2}$.
Retardation is the magnitude of negative acceleration,so retardation = $12.5 \text{ m s}^{-2}$.
(ii) To find time $(t)$: Using the first equation of motion,$v = u + at$.
$0 = 25 + (-12.5) \times t$.
$12.5t = 25$.
$t = \frac{25}{12.5} = 2 \text{ s}$.

(A) Given: Acceleration $a = -6 \, m s^{-2}$ (negative because it is in the opposite direction of motion),time $t = 2 \, s$,and final velocity $v = 0 \, m s^{-1}$ (since the car stops).
$(i)$ First,find the initial velocity $u$ using the first equation of motion: $v = u + at$.
$0 = u + (-6) \times 2$
$u = 12 \, m s^{-1}$.
$(ii)$ Now,calculate the distance $S$ using the second equation of motion: $S = ut + \frac{1}{2}at^2$.
$S = (12 \times 2) + \frac{1}{2} \times (-6) \times (2)^2$
$S = 24 - 12 = 12 \, m$.
Therefore,the car travels $12 \, m$ before stopping.

(A) Given: Initial velocity $u = 36 \, km \, h^{-1} = 36 \times \frac{5}{18} = 10 \, m \, s^{-1}$.
Final velocity $v = 0 \, m \, s^{-1}$ (since the cycle stops).
Time $t = 2 \, s$.
$(i)$ To calculate retardation:
Using the first equation of motion,$v = u + at$:
$0 = 10 + a \times 2$
$2a = -10$
$a = -5 \, m \, s^{-2}$.
Retardation is the negative of acceleration,so retardation $= -(-5 \, m \, s^{-2}) = 5 \, m \, s^{-2}$.
$(ii)$ To calculate distance covered $(S)$:
Using the third equation of motion,$v^2 - u^2 = 2aS$:
$0^2 - (10)^2 = 2 \times (-5) \times S$
$-100 = -10 \times S$
$S = \frac{-100}{-10} = 10 \, m$.

(N/A) $(i)$ Given: Initial velocity $u = 0 \, m/s$,Distance $S = 64 \, m$,Time $t = 4 \, s$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$:
$64 = 0(4) + \frac{1}{2} \times a \times (4)^2$
$64 = 8a$
$a = 8 \, m/s^2$.
Using the equation $v = u + at$:
$v = 0 + 8 \times 4 = 32 \, m/s$.
$(ii)$ Half the total distance is $S' = \frac{64}{2} = 32 \, m$.
Using $S' = ut + \frac{1}{2}at^2$:
$32 = 0(t) + \frac{1}{2} \times 8 \times t^2$
$32 = 4t^2$
$t^2 = 8$
$t = \sqrt{8} \approx 2.83 \, s$.

(C) For a body starting from rest $(u = 0)$ and moving with constant acceleration $(a)$ along an inclined plane, the distance $(s)$ covered in time $(t)$ is given by the equation of motion: $s = ut + \frac{1}{2}at^2$.
Since $u = 0$, the equation simplifies to $s = \frac{1}{2}at^2$, which implies $s \propto t^2$.
Let the total distance be $S$ and the total time taken be $T = 4$ seconds.
We are asked to find the time $(t')$ taken to cover one-fourth of the distance, i.e., $s' = \frac{S}{4}$.
Using the proportionality $s \propto t^2$, we have the ratio: $\frac{s'}{S} = \frac{(t')^2}{T^2}$.
Substituting the values: $\frac{S/4}{S} = \frac{(t')^2}{4^2}$.
$\frac{1}{4} = \frac{(t')^2}{16}$.
$(t')^2 = \frac{16}{4} = 4$.
$t' = \sqrt{4} = 2$ seconds.
Therefore, the body covers one-fourth of the distance in $2$ seconds.

(A) Let the velocity at point $A$ be $v_A = 20 \, m/s$ and at point $B$ be $v_B = 30 \, m/s$.
Let the distance between $A$ and $B$ be $s$. The velocity $v$ at the midpoint $M$ can be found using the third equation of motion: $v^2 = u^2 + 2as$.
For the segment $AM$,$v_M^2 = v_A^2 + 2a(s/2) = v_A^2 + as$.
For the segment $MB$,$v_B^2 = v_M^2 + 2a(s/2) = v_M^2 + as$.
From the first equation,$as = v_M^2 - v_A^2$. Substituting this into the second equation:
$v_B^2 = v_M^2 + (v_M^2 - v_A^2) = 2v_M^2 - v_A^2$.
Rearranging for $v_M$:
$2v_M^2 = v_A^2 + v_B^2 \implies v_M = \sqrt{\frac{v_A^2 + v_B^2}{2}}$.
Substituting the values: $v_M = \sqrt{\frac{20^2 + 30^2}{2}} = \sqrt{\frac{400 + 900}{2}} = \sqrt{\frac{1300}{2}} = \sqrt{650} = \sqrt{25 \times 26} = 5 \sqrt{26} \approx 25.49 \, m/s$.
Wait,calculating $\sqrt{650} \approx 25.495$. Let's re-check the options. $10\sqrt{6} = \sqrt{600} \approx 24.49$. Actually,the formula for velocity at the midpoint is $v_M = \sqrt{\frac{v_A^2 + v_B^2}{2}}$. Calculation: $\sqrt{650} \approx 25.5 \, m/s$.

(N/A) For a body thrown vertically upwards with initial velocity $u$,the final velocity at the maximum height is $v = 0$.
Let $t_{1}$ be the time of ascent. Using the first equation of motion $v = u + at$,where $a = -g$:
$0 = u - gt_{1}$
$\Rightarrow t_{1} = \frac{u}{g} ......(1)$
For the return journey (descent),the body starts from rest at the maximum height,so initial velocity $u' = 0$. The final velocity when it reaches the ground is $v' = u$. Let $t_{2}$ be the time of descent. Using $v' = u' + gt_{2}$:
$u = 0 + gt_{2}$
$\Rightarrow t_{2} = \frac{u}{g} ......(2)$
Comparing equation $(1)$ and $(2)$,we get $t_{1} = t_{2}$.
Thus,the time of ascent is equal to the time of descent.

$(15 M S^{-1})$ For the first ball:
$u = 0, h = 20\, m, g = 10\, m s^{-2}, t = ?$
Using the equation of motion $S = ut + \frac{1}{2}at^2$, we have:
$20 = 0 + \frac{1}{2} \times 10 \times t^2$
$20 = 5t^2$
$t^2 = 4$
$t = 2\, s$
For the second ball:
Since the second ball is thrown $1\, s$ later and reaches the ground at the same time, its time of travel is $t' = 2 - 1 = 1\, s$.
$u = ?, h = 20\, m, g = 10\, m s^{-2}, t' = 1\, s$
Using $S = ut' + \frac{1}{2}at'^2$, we have:
$20 = u(1) + \frac{1}{2} \times 10 \times (1)^2$
$20 = u + 5$
$u = 20 - 5 = 15\, m s^{-1}$

(A) Given:
Initial velocity $(u)$ = $0 \, m s^{-1}$ (as the stone is dropped).
Distance fallen $(s)$ = $20 \, m$.
Acceleration due to gravity $(g)$ = $10 \, m s^{-2}$.
Using the third equation of motion: $v^2 - u^2 = 2as$.
Substituting the values: $v^2 - 0^2 = 2 \times 10 \times 20$.
$v^2 = 400$.
$v = \sqrt{400} = 20 \, m s^{-1}$.
Thus,the speed of the stone after falling $20 \, m$ is $20 \, m s^{-1}$.

(B) To find the initial velocity $(u)$ of the ball,we use the third equation of motion: $v^2 = u^2 + 2as$.
At maximum height,the final velocity $(v)$ is $0 \, m/s$.
The acceleration due to gravity $(a)$ is $-g = -9.8 \, m/s^2$.
The displacement $(s)$ is $100 \, m$.
Substituting these values into the equation: $0^2 = u^2 + 2(-9.8)(100)$.
$0 = u^2 - 1960$.
$u^2 = 1960$.
$u = \sqrt{1960} \approx 44.27 \, m/s$.
Rounding to the nearest option,the speed is $44.2 \, m/s$.