Motion Under Gravity Questions in English

(C) Let the height of point $Q$ be $h$ and the height of point $P$ be $2h$. Let the time taken to reach the ground be $t$. Taking the downward direction as negative and the ground as the origin $(y=0)$:
For particle at $P$: Initial velocity $u_P = -5 \, m/s$,displacement $s_P = -2h$. Using $s = ut + \frac{1}{2}at^2$,we get $-2h = -5t - 5t^2$,which simplifies to $2h = 5t + 5t^2$ (Equation $1$).
For particle at $Q$: Initial velocity $u_Q = +5 \, m/s$,displacement $s_Q = -h$. Using $s = ut + \frac{1}{2}at^2$,we get $-h = 5t - 5t^2$,which simplifies to $h = 5t^2 - 5t$ (Equation $2$).
Substitute $h$ from Equation $2$ into Equation $1$: $2(5t^2 - 5t) = 5t + 5t^2 \Rightarrow 10t^2 - 10t = 5t + 5t^2 \Rightarrow 5t^2 = 15t$. Since $t \neq 0$,we get $t = 3 \, s$.
Now,substitute $t = 3$ into Equation $2$: $h = 5(3)^2 - 5(3) = 45 - 15 = 30 \, m$.
The distance between $P$ and $Q$ is $PQ = 2h - h = h = 30 \, m$. Thus,both $(a)$ and $(b)$ are correct.

(C) The juggler maintains $4$ balls in the air. The time interval between throwing consecutive balls is $T = 1/4 \, s$.
Since there are $4$ balls,the total time a ball stays in the air (time of flight) must be $T_{flight} = 4 \times T = 4 \times (1/4) = 1 \, s$.
The time taken to reach the maximum height is $t = T_{flight} / 2 = 1 / 2 = 0.5 \, s$.
Using the equation of motion for a ball thrown vertically upward,at maximum height,the final velocity $v = 0$.
Using $v = u - gt$,we get $0 = u - (10)(0.5)$,so $u = 5 \, m/s$.
The maximum height $H$ reached is given by $H = \frac{u^2}{2g} = \frac{5^2}{2 \times 10} = \frac{25}{20} = 1.25 \, m$.
Alternatively,using $H = \frac{1}{2}gt^2$ for the downward journey from the peak: $H = \frac{1}{2} \times 10 \times (0.5)^2 = 5 \times 0.25 = 1.25 \, m$.

(C) Let $H$ be the total height of the tower and $t$ be the total time taken to reach the ground.
Using the equation of motion for a body dropped from rest: $H = \frac{1}{2} gt^2$ --- $(1)$
In the last second,the body covers $7/16$ of the total height,meaning it covers $H - \frac{7}{16}H = \frac{9}{16}H$ in the first $(t-1)$ seconds.
So,$\frac{9}{16}H = \frac{1}{2} g(t-1)^2$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{H}{\frac{9}{16}H} = \frac{\frac{1}{2} gt^2}{\frac{1}{2} g(t-1)^2}$
$\frac{16}{9} = \frac{t^2}{(t-1)^2}$
Taking the square root on both sides:
$\frac{4}{3} = \frac{t}{t-1}$
$4(t-1) = 3t$
$4t - 4 = 3t$
$t = 4 \ s$

(C) $1$. At the instant the stone is released,it possesses the same velocity as the balloon due to inertia. Therefore,the velocity of the stone immediately after release is $v$ (upward).
$2$. Once the stone is released,it is no longer in contact with the balloon and is only under the influence of gravity.
$3$. The only force acting on the stone is its weight,which acts vertically downward.
$4$. According to Newton's second law,the acceleration of the stone is $F/m = mg/m = g$.
$5$. Thus,the acceleration of the stone is $g$ (downward).

(A) Let the total time of motion be $n$ seconds. The initial velocity $u = 0$ and acceleration $g = 10\,m/s^2$.
Distance covered in the first $3$ seconds is given by $S_3 = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2} \times 10 \times 3^2 = 45\,m$.
Distance covered in the last second of motion is given by $S_{last} = u + \frac{g}{2}(2n - 1) = 0 + \frac{10}{2}(2n - 1) = 5(2n - 1)$.
According to the problem,$S_{last} = S_3$,so $5(2n - 1) = 45$.
$2n - 1 = 9 \Rightarrow 2n = 10 \Rightarrow n = 5\,s$.
The total height $h$ is the distance covered in $5$ seconds: $h = \frac{1}{2}gn^2 = \frac{1}{2} \times 10 \times 5^2 = 5 \times 25 = 125\,m$.

(D) When objects are dropped from the same height,they undergo free fall under the influence of gravity.
According to the equations of motion,the acceleration of an object in free fall is independent of its mass and is equal to the acceleration due to gravity,$g$.
Since both spheres are dropped from the same height and are at the same position ($1\, m$ above the ground),they have the same velocity.
However,momentum $(p = mv)$,kinetic energy $(K = \frac{1}{2}mv^2)$,and potential energy $(U = mgh)$ all depend on the mass of the object.
Since the masses are different ($2\, kg$ and $4\, kg$),their momentum,kinetic energy,and potential energy will be different.
Therefore,the only quantity that remains the same for both spheres is their acceleration,which is $g$.

(C) For a particle moving under gravity,the acceleration is always equal to the acceleration due to gravity $(g)$,which is approximately $9.8\,ms^{-2}$ directed downwards.
Since the acceleration is constant throughout the motion,the acceleration at $t = 1\,s$ is $a_1 = 9.8\,ms^{-2}$.
Similarly,the acceleration at $t = 2\,s$ is $a_2 = 9.8\,ms^{-2}$.
The ratio of the accelerations is $\frac{a_1}{a_2} = \frac{9.8}{9.8} = 1$.

(C) The time taken by the body to reach the highest point is given by the equation of motion $v = u - gt$.
At the highest point,the final velocity $v = 0$.
Given $u = 19.6 \, ms^{-1}$ and $g = 9.8 \, ms^{-2}$.
$0 = 19.6 - 9.8 \times t$
$t = \frac{19.6}{9.8} = 2 \, s$.
Since the time taken to go up is equal to the time taken to come down,the body will return to the starting point in $2 \, s + 2 \, s = 4 \, s$.

(B) The equation of motion for height is $y(t) = ut - \frac{1}{2}gt^2$.
From the graph,the ball is at the same height at $t_1 = 1\,s$ and $t_4 = 6\,s$. The time of flight is $T = t_1 + t_4 = 1 + 6 = 7\,s$.
The time to reach the maximum height is $t_{max} = \frac{T}{2} = 3.5\,s$.
At $t_{max} = 3.5\,s$,the velocity is zero,so $u = gt_{max} = 7.5 \times 3.5 = 26.25\,m/s$.
The height at $t = 2\,s$ is $y(2) = u(2) - \frac{1}{2}g(2)^2 = 26.25(2) - 0.5(7.5)(4) = 52.5 - 15 = 37.5\,m$.
The height at $t = 1\,s$ is $y(1) = u(1) - \frac{1}{2}g(1)^2 = 26.25(1) - 0.5(7.5)(1) = 26.25 - 3.75 = 22.5\,m$.
The height $h$ is the difference between the height at $t = 2\,s$ and $t = 1\,s$: $h = y(2) - y(1) = 37.5 - 22.5 = 15\,m$.

(A) The total height $h$ of the tower is given by the equation of motion for a body falling from rest: $h = \frac{1}{2}gT^2$.
At time $t = T/3$,the distance $h'$ covered by the ball from the top is: $h' = \frac{1}{2}g(T/3)^2 = \frac{1}{2}g(T^2/9) = \frac{1}{9}(\frac{1}{2}gT^2) = \frac{h}{9}$.
The position of the ball from the ground is the total height minus the distance covered from the top: $h_{ground} = h - h' = h - \frac{h}{9} = \frac{8h}{9}$ meters.

(C) Let the first body fall for time $n$. Its displacement is $y_1 = \frac{1}{2} g n^2$.
The second body starts after a time interval $N$,so it falls for time $(n - N)$. Its displacement is $y_2 = \frac{1}{2} g (n - N)^2$.
The vertical separation between the two bodies is given as $1 \, m$,so $y_1 - y_2 = 1$.
Substituting the expressions: $\frac{1}{2} g n^2 - \frac{1}{2} g (n - N)^2 = 1$.
Expanding the term $(n - N)^2 = n^2 - 2nN + N^2$:
$\frac{g}{2} [n^2 - (n^2 - 2nN + N^2)] = 1$.
$\frac{g}{2} [2nN - N^2] = 1$.
$gNn - \frac{gN^2}{2} = 1$.
$gNn = 1 + \frac{gN^2}{2}$.
Dividing by $gN$: $n = \frac{1}{gN} + \frac{N}{2}$.

(A) The initial velocity of the packet is the same as the velocity of the balloon,so $u = 12\,ms^{-1}$.
Since the packet is released from a height,its displacement when it reaches the ground is $s = -65\,m$ (taking the upward direction as positive).
The acceleration due to gravity is $g = -10\,ms^{-2}$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$-65 = 12t + \frac{1}{2}(-10)t^2$
$-65 = 12t - 5t^2$
$5t^2 - 12t - 65 = 0$
Solving the quadratic equation using the formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$t = \frac{12 \pm \sqrt{(-12)^2 - 4(5)(-65)}}{2(5)}$
$t = \frac{12 \pm \sqrt{144 + 1300}}{10}$
$t = \frac{12 \pm \sqrt{1444}}{10}$
$t = \frac{12 \pm 38}{10}$
Since time cannot be negative,we take $t = \frac{12 + 38}{10} = \frac{50}{10} = 5\,s$.

(A) When a body is thrown vertically upwards,at the highest point,its velocity becomes $0 \ m/s$.
However,the gravitational force continues to act on it,resulting in a constant downward acceleration of $g \approx 9.8 \ m/s^2$.
Thus,a body can have acceleration even when its velocity is zero at a specific instant.
Therefore,the $Assertion$ is correct.
When a body reverses its direction of motion,it must momentarily come to rest (velocity becomes zero) before moving in the opposite direction.
This change in velocity is caused by the acceleration acting on the body.
Thus,the $Reason$ is correct and it provides a valid explanation for why the velocity can be zero while acceleration is non-zero.

$(A)$ When an object is thrown vertically upward with an initial velocity $u$, it reaches a maximum height $H = \frac{u^2}{2g}$.
Since the maximum height depends only on the initial velocity $u$ and acceleration due to gravity $g$, it is independent of the mass $m$ of the object.
According to the law of conservation of energy, the kinetic energy at the point of projection is $K = \frac{1}{2}mu^2$. When the object returns to the same point, its potential energy is the same as at the start, so its kinetic energy must be the same, meaning its speed $v = u$.
Thus, both the maximum height and the final speed at the point of projection are independent of the mass of the ball.
Therefore, both the $Assertion$ and $Reason$ are correct, and the $Reason$ is the correct explanation of the $Assertion$.

(A) For the upward motion,the final velocity at the highest point is $v = 0$.
Using the equation of motion $v = u + at$,for the last second of motion,the initial velocity $u'$ at the start of that last second is $u' = v - at = 0 - (-g)(1) = g$.
The distance covered in the last second is $s = u't + \frac{1}{2}at^2 = g(1) + \frac{1}{2}(-g)(1)^2 = g - \frac{g}{2} = \frac{g}{2}$.
Taking $g \approx 10 \ m/s^2$,we get $s = \frac{10}{2} = 5 \ m$.
This distance is independent of the initial velocity of projection.
The reason is also correct because the motion is symmetric; the distance covered in the last second of upward motion is identical to the distance covered in the first second of free fall (downward motion) starting from rest $(u=0)$.
Thus,both Assertion and Reason are correct,and the Reason explains the Assertion.

(B) Using the third equation of motion: $v^2 - u^2 = 2as$.
Here,initial velocity $u = 0 \ m/s$ (as water falls from rest),acceleration $a = g = 9.8 \ m/s^2$,and displacement $s = 19.6 \ m$.
Substituting the values:
$v^2 - 0^2 = 2 \times 9.8 \times 19.6$
$v^2 = 2 \times 9.8 \times (2 \times 9.8)$
$v^2 = (2 \times 9.8)^2$
$v = 2 \times 9.8 = 19.6 \ m/s$.
Therefore,the velocity of water at the turbines is $19.6 \ m/s$.

(D) The time taken for an object to fall a distance $h$ under constant acceleration $a$ is given by $t = \sqrt{\frac{2h}{a}}$.
When the elevator is at rest,its acceleration is $0$,so the effective acceleration of the coin relative to the floor is $g$.
When the elevator is moving uniformly (constant velocity),its acceleration is also $0$. Therefore,the effective acceleration of the coin relative to the floor remains $g$.
Since the effective acceleration is the same in both cases $(a_{real} = g)$,the time taken for the coin to reach the floor is the same.
Thus,$t_{1} = t_{2}$.

(B) Let the total time taken to reach the ground be $T$ seconds.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ (dropped from rest):
For the total distance $100\; m$,we have $100 = \frac{1}{2}aT^2 \implies T = \sqrt{\frac{200}{a}}$.
In the last $\frac{1}{2}\; s$,the ball covers $19\; m$. This means in time $(T - 0.5)\; s$,the ball covers $(100 - 19) = 81\; m$.
So,$81 = \frac{1}{2}a(T - 0.5)^2 \implies T - 0.5 = \sqrt{\frac{162}{a}}$.
Substituting $T = \sqrt{\frac{200}{a}}$ into the equation:
$\sqrt{\frac{200}{a}} - 0.5 = \sqrt{\frac{162}{a}}$
$\frac{10\sqrt{2}}{\sqrt{a}} - \frac{9\sqrt{2}}{\sqrt{a}} = 0.5$
$\frac{\sqrt{2}}{\sqrt{a}} = 0.5$
$\sqrt{\frac{2}{a}} = \frac{1}{2}$
Squaring both sides: $\frac{2}{a} = \frac{1}{4} \implies a = 8\; m/s^2$.

(B) Let the initial velocity of the ball be $v_0 = 20 \; m s^{-1}$ in the upward direction.
At the maximum height,the final velocity of the ball is $v = 0 \; m s^{-1}$.
The acceleration due to gravity is $a = -g = -10 \; m s^{-2}$.
Let $h$ be the maximum height reached by the ball above the point of projection.
Using the kinematic equation $v^2 = v_0^2 + 2ah$,we have:
$0^2 = (20)^2 + 2(-10)h$
$0 = 400 - 20h$
$20h = 400$
$h = 20 \; m$.
Therefore,the ball will rise $20 \; m$ above the point of projection.

(C) Let the ground be the origin $(y = 0)$ and the vertically upward direction be positive.
Initial position $y_0 = 25 \; m$,initial velocity $v_0 = 20 \; m s^{-1}$,and acceleration $a = -g = -10 \; m s^{-2}$.
The position of the ball at any time $t$ is given by the kinematic equation:
$y = y_0 + v_0 t + \frac{1}{2} a t^2$
When the ball hits the ground,$y = 0$. Substituting the values:
$0 = 25 + 20t + \frac{1}{2}(-10)t^2$
$0 = 25 + 20t - 5t^2$
Dividing by $-5$:
$t^2 - 4t - 5 = 0$
Factoring the quadratic equation:
$(t - 5)(t + 1) = 0$
Since time cannot be negative,we take $t = 5 \; s$.

(N/A) An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by $g$. If air resistance is neglected,the object is said to be in free fall. If the height through which the object falls is small compared to the Earth's radius,$g$ can be taken to be constant,equal to $9.8 \ m \ s^{-2}$. Free fall is thus a case of motion with uniform acceleration.
We assume that the motion is in $y$-direction,more correctly in $-y$-direction because we choose the upward direction as positive. Since the acceleration due to gravity is always downward,it is in the negative direction and we have $a = -g = -9.8 \ m \ s^{-2}$.
The object is released from rest at $y = 0$. Therefore,$v_0 = 0$ and the equations of motion become:
$v = 0 - gt = -9.8t \ m \ s^{-1}$
$y = 0 - 1/2 gt^2 = -4.9t^2 \ m$
$v^2 = 0 - 2gy = -19.6y \ m^2 \ s^{-2}$
These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance. The variation of acceleration,velocity,and distance with time are shown in the figures.

(N/A) Let us divide the time interval of motion of an object under free fall into many equal intervals $\tau$ and find out the distances traversed during successive intervals of time. Since the initial velocity is zero, the position $y$ at time $t$ is given by:
$y = -\frac{1}{2} g t^2$
Using this equation, we calculate the position of the object after time intervals $t = 0, \tau, 2\tau, 3\tau, \dots$. If we define $y_0 = -\frac{1}{2} g \tau^2$ as the position after the first interval $\tau$, then the position at time $n\tau$ is $n^2 y_0$. The distance traversed in the $n^{th}$ interval is the difference between the position at $n\tau$ and $(n-1)\tau$:
Distance in $n^{th}$ interval $= |n^2 y_0 - (n-1)^2 y_0| = |(n^2 - (n^2 - 2n + 1)) y_0| = (2n - 1) |y_0|$.
For $n = 1, 2, 3, 4, \dots$, the distances are $1|y_0|, 3|y_0|, 5|y_0|, 7|y_0|, \dots$. Thus, the ratio of distances is $1: 3: 5: 7: \dots$, which are odd numbers. This law was established by Galileo Galilei ($1564$-$1642$), who was the first to make quantitative studies of free fall.

(C) The ruler drops under free fall. Therefore,the initial velocity $u = 0$ and acceleration $a = g = 9.8 \; m/s^2$.
The distance travelled $d$ and the reaction time $t_r$ are related by the kinematic equation:
$d = ut_r + \frac{1}{2}gt_r^2$
Since $u = 0$,we have:
$d = \frac{1}{2}gt_r^2$
$t_r = \sqrt{\frac{2d}{g}}$
Given $d = 21.0 \; cm = 0.21 \; m$ and $g = 9.8 \; m/s^2$:
$t_r = \sqrt{\frac{2 \times 0.21}{9.8}}$
$t_r = \sqrt{\frac{0.42}{9.8}}$
$t_r = \sqrt{0.0428} \approx 0.207 \; s$
Rounding to one significant figure or the nearest given option,the reaction time is approximately $0.2 \; s$.

(B) The acceleration acting on the ball is the acceleration due to gravity $(g)$.
Irrespective of the direction of the motion of the ball (whether it is moving upwards or downwards),the acceleration due to gravity always acts vertically downwards towards the center of the Earth.
Therefore,the direction of acceleration is downwards.

(A) At the maximum height,the vertical velocity of the ball becomes zero momentarily as it changes direction from upward to downward.
Acceleration due to gravity $(g)$ is a constant force acting on the object throughout its flight,regardless of its position or velocity.
Therefore,at the highest point,the velocity is $0\; m s^{-1}$ and the acceleration is $9.8\; m s^{-2}$ directed downwards.

(N/A) Given that the highest point is $x=0$ and the downward direction is positive:
$1$. During upward motion: The ball is above the highest point (which is $x=0$),so the position $x$ is negative. The ball is moving upwards (opposite to the positive $x$-axis),so the velocity $v$ is negative. The acceleration due to gravity $g$ always acts downwards,which is the positive direction,so the acceleration $a$ is positive.
$2$. During downward motion: The ball is below the highest point,so the position $x$ is positive. The ball is moving downwards (along the positive $x$-axis),so the velocity $v$ is positive. The acceleration due to gravity $g$ acts downwards,which is the positive direction,so the acceleration $a$ is positive.

(C) Given: Initial velocity $u = 29.4 \; m s^{-1}$,final velocity at maximum height $v = 0 \; m s^{-1}$,acceleration $a = -g = -9.8 \; m s^{-2}$.
Using the third equation of motion,$v^{2} - u^{2} = 2as$:
$0^{2} - (29.4)^{2} = 2(-9.8)s$
$s = \frac{-(29.4)^{2}}{-19.6} = \frac{864.36}{19.6} = 44.1 \; m$.
Using the first equation of motion,$v = u + at$ to find the time of ascent $(t_{a})$:
$0 = 29.4 + (-9.8)t_{a}$
$t_{a} = \frac{-29.4}{-9.8} = 3 \; s$.
Since the time of ascent equals the time of descent,the total time taken to return to the player's hands is $T = t_{a} + t_{d} = 3 + 3 = 6 \; s$.

(N/A) The ball is dropped from a height $s = 90 \; m$.
Initial velocity $u = 0$,acceleration $a = g = 9.8 \; m/s^2$.
Time $t$ taken to hit the ground: $s = ut + (1/2)at^2 \implies 90 = 0 + (1/2) \times 9.8 \times t^2 \implies t = \sqrt{18.367} \approx 4.29 \; s$.
Final velocity just before the first impact: $v = u + at = 0 + 9.8 \times 4.29 = 42.04 \; m/s$.
Rebound velocity $u_r = (9/10)v = 0.9 \times 42.04 = 37.84 \; m/s$.
Time to reach maximum height after first rebound: $v = u_r + at' \implies 0 = 37.84 - 9.8 \times t' \implies t' = 3.86 \; s$.
Total time for the ball to return to the floor: $t + 2t' = 4.29 + 2 \times 3.86 = 12.01 \; s$.
Velocity just before the second impact: $v_2 = 37.84 - 9.8 \times 3.86 = 0 \; m/s$ (at peak),then it accelerates back down to $37.84 \; m/s$ at the floor.
After the second collision,the new rebound velocity is $u_{r2} = (9/10) \times 37.84 = 34.06 \; m/s$.

(N/A) No,it is not correct. The $x-t$ graph of a particle moving in a straight line for $t < 0$ and on a parabolic path for $t > 0$ cannot be represented by the given graph.
This is because the graph shows the particle at rest $(x = 0)$ for $t < 0$,and then it starts moving with acceleration for $t > 0$.
$A$ suitable physical context for this graph is a ball held at a height $x = 0$ and released at $t = 0$ to fall freely under gravity.

(N/A) In the given $v-t$ graph,the sign of velocity changes periodically,and its magnitude decreases with the passage of time.
This indicates that the object is undergoing motion with constant acceleration (due to gravity) but loses a fraction of its velocity upon each collision.
$A$ suitable physical situation for this graph is a ball dropped from a height onto a hard floor.
When the ball strikes the floor,it rebounds with a velocity smaller than the velocity with which it struck the floor,due to energy loss during the collision.
This process repeats with each subsequent bounce,leading to a decrease in the peak velocity and the time interval between bounces,until the ball eventually comes to rest.

(10 S, 10 S) Initial velocity of the ball,$u = 49\; m s^{-1}$.
Acceleration due to gravity,$a = -g = -9.8\; m s^{-2}$.
Case $I$: When the lift is stationary.
Taking the upward motion of the ball,the final velocity $v$ at the highest point is $0$.
Using the first equation of motion,$v = u + at$,the time of ascent $t_a$ is:
$t_a = \frac{v - u}{a} = \frac{0 - 49}{-9.8} = 5\; s$.
Since the time of ascent equals the time of descent,the total time taken to return to the hand is $T = t_a + t_d = 5 + 5 = 10\; s$.
Case $II$: When the lift moves up with a uniform velocity of $5\; m s^{-1}$.
Since the lift moves with a uniform velocity,its acceleration is $0$. The relative velocity of the ball with respect to the boy remains $49\; m s^{-1}$ (the same as in the stationary case). Because the frame of reference (the lift) is inertial,the time taken for the ball to return to the boy's hand remains the same,which is $10\; s$.

(N/A) For the first stone:
Initial velocity,$u_{1} = 15 \; m s^{-1}$
Acceleration,$a = -g = -10 \; m s^{-2}$
Using the equation of motion,$x_{1} = x_{0} + u_{1}t + \frac{1}{2}at^{2}$
Given the height of the cliff,$x_{0} = 200 \; m$,so $x_{1} = 200 + 15t - 5t^{2} \; \dots (i)$
When this stone hits the ground,$x_{1} = 0$,so $-5t^{2} + 15t + 200 = 0 \implies t^{2} - 3t - 40 = 0$.
Solving the quadratic equation,$(t - 8)(t + 5) = 0$. Since $t > 0$,$t = 8 \; s$.
For the second stone:
Initial velocity,$u_{2} = 30 \; m s^{-1}$
Acceleration,$a = -g = -10 \; m s^{-2}$
$x_{2} = 200 + 30t - 5t^{2} \; \dots (ii)$
When this stone hits the ground,$x_{2} = 0$,so $-5t^{2} + 30t + 200 = 0 \implies t^{2} - 6t - 40 = 0$.
Solving the quadratic equation,$(t - 10)(t + 4) = 0$. Since $t > 0$,$t = 10 \; s$.
For $0 \le t \le 8 \; s$,both stones are in the air:
$x_{2} - x_{1} = (200 + 30t - 5t^{2}) - (200 + 15t - 5t^{2}) = 15t$.
This is a linear equation representing the straight line part of the graph.
For $8 \; s < t \le 10 \; s$,only the second stone is in the air $(x_{1} = 0)$:
$x_{2} - x_{1} = x_{2} - 0 = 200 + 30t - 5t^{2}$.
This is a quadratic equation representing the curved part of the graph.

(A) The total time taken to hear the splash is the sum of the time taken by the stone to fall $(t_1)$ and the time taken by the sound to travel back to the top $(t_2)$.
$1$. Time taken by the stone to reach the water $(t_1)$:
Using the equation of motion $s = ut + \frac{1}{2}gt^2$,where $u = 0$,$s = 300 \; m$,and $g = 9.8 \; m s^{-2}$:
$300 = 0 + \frac{1}{2} \times 9.8 \times t_1^2$
$t_1^2 = \frac{600}{9.8} \approx 61.22$
$t_1 = \sqrt{61.22} \approx 7.82 \; s$
$2$. Time taken by the sound to reach the top $(t_2)$:
Using $t_2 = \frac{\text{distance}}{\text{speed}}$:
$t_2 = \frac{300}{340} \approx 0.88 \; s$
$3$. Total time $(t)$:
$t = t_1 + t_2 = 7.82 + 0.88 = 8.7 \; s$.

(N/A) Galileo concluded that for all objects in free fall,the rate of change of velocity with respect to time (acceleration) is a constant of motion,which is independent of the mass or shape of the object.
Conversely,he observed that the change in velocity with respect to distance is not constant.
Specifically,the rate of change of velocity with distance decreases as the distance of the fall increases.

(N/A) Galileo Galilei challenged the Aristotelian view that heavier objects fall faster than lighter ones. Through his experiments and logical reasoning,he observed that in the absence of air resistance,all objects fall towards the Earth with the same constant acceleration,regardless of their mass. This acceleration is known as the acceleration due to gravity,denoted by $g$,which is approximately $9.8 \ m/s^2$ near the Earth's surface. Galileo concluded that if two objects of different masses are dropped simultaneously from the same height in a vacuum,they will reach the ground at the same time.

(N/A) The motion of an object falling solely under the influence of gravitational force is called free fall.
In this motion,the acceleration produced is the acceleration due to gravity $(g)$,which acts in the downward direction.
If air resistance is neglected,the motion is considered uniformly accelerated with acceleration $a = -g$.
By substituting the initial velocity $v_{0} = 0$,acceleration $a = -g$,and displacement $d = -h$ into the standard kinematic equations,we get:
$1$. For velocity: $v = v_{0} + at \implies v = -gt$
$2$. For displacement: $d = v_{0}t + \frac{1}{2}at^{2} \implies -h = 0 - \frac{1}{2}gt^{2} \implies h = \frac{1}{2}gt^{2}$
$3$. For velocity-displacement relation: $v^{2} - v_{0}^{2} = 2ad \implies v^{2} - 0 = 2(-g)(-h) \implies v^{2} = 2gh \implies v = \sqrt{2gh}$

(N/A) For an object falling freely,the acceleration is constant and equal to $-g$. Thus,graph $(a)$ shows the constant acceleration with respect to time $(a \to t)$.
The velocity of an object falling freely is given by $v = -gt$. This represents a linear relationship with time,as shown in graph $(b)$ for $v \to t$.
The displacement of an object falling freely is given by $x = -\frac{1}{2}gt^2$ (taking downward as negative). This represents a parabolic relationship with time,as shown in graph $(c)$ for $x \to t$.

(A) An object is said to be in free fall when it is moving solely under the influence of gravity,without any initial push or throw.
By definition,for an object falling freely from rest,the initial velocity $(u)$ is $0 \ m/s$.

(N/A) For an object falling freely under gravity,the initial velocity $u = 0$ and acceleration $a = g$.
The distance covered by an object in the $n^{th}$ second is given by the formula:
$S_n = u + \frac{a}{2}(2n - 1)$
Substituting $u = 0$ and $a = g$ into the equation:
$S_n = 0 + \frac{g}{2}(2n - 1)$
Therefore,the distance covered in the $n^{th}$ second is $S_n = \frac{g}{2}(2n - 1)$.

(N/A) For an object falling freely from a height $h$ under gravity,the initial velocity $u$ is $0 \ m/s$.
Using the third equation of motion: $v^2 - u^2 = 2as$.
Here,$v$ is the final velocity,$u = 0$,$a = g$ (acceleration due to gravity),and $s = h$ (displacement).
Substituting these values: $v^2 - 0^2 = 2gh$.
Therefore,the equation for velocity is $v = \sqrt{2gh}$.

(N/A) For an object falling freely under gravity starting from rest,the equation of motion is given by $y = \frac{1}{2}gt^2$,where $g$ is the acceleration due to gravity.
Since $y$ is proportional to the square of time $(y \propto t^2)$,the graph of $y$ versus $t$ is a parabola opening upwards.
Starting from the origin $(0,0)$,the curve increases as $t$ increases,representing the increasing displacement of the object as it accelerates downwards.

(N/A) Yes,a particle moving in a straight line can have zero velocity and non-zero acceleration at the same instant.
For example,when a ball is thrown vertically upwards,at the highest point of its trajectory,its instantaneous velocity is $0 \ m/s$,but it still experiences a constant acceleration due to gravity $(g \approx 9.8 \ m/s^2)$ acting downwards.

(B) For a body falling freely from rest,the initial velocity $u = 0$ and acceleration $a = g$.
The distance covered in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Substituting $u = 0$ and $a = g$,we get $S_n = \frac{g}{2}(2n - 1)$.
For $n = 1$: $S_1 = \frac{g}{2}(2(1) - 1) = \frac{g}{2}(1) = \frac{g}{2}$.
For $n = 2$: $S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3) = \frac{3g}{2}$.
For $n = 3$: $S_3 = \frac{g}{2}(2(3) - 1) = \frac{g}{2}(5) = \frac{5g}{2}$.
The ratio of distances $S_1 : S_2 : S_3$ is $\frac{g}{2} : \frac{3g}{2} : \frac{5g}{2} = 1 : 3 : 5$.

(A) For a freely falling body,the initial velocity $u = 0 \ m/s$ and acceleration $a = g = 10 \ m/s^2$.
Using the kinematic equation for distance covered in time $t$:
$s = ut + \frac{1}{2}at^2$
Substituting the values for $t = 1 \ s$:
$s = (0)(1) + \frac{1}{2}(10)(1)^2$
$s = 0 + 5(1)$
$s = 5 \ m$
Therefore,the distance covered in the first second is $5 \ m$.

(N/A) It is clear from the graph that displacement $x$ is positive during motion. The ball is dropped from a height and its velocity increases in the downward direction due to the gravitational field. In this condition,$v$ is negative,but the acceleration of the ball is equal to the acceleration due to gravity,i.e.,$a = -g$.
When the ball rebounds in the upward direction,its velocity is positive,but the acceleration remains $a = -g$.
$(a)$ The velocity-time graph of the ball is shown in fig. $(i)$.
$(b)$ The acceleration-time graph of the ball is shown in fig. $(ii)$.

(N/A) When a football is kicked vertically upwards,it moves under the influence of gravity alone (ignoring air resistance).
$(a)$ At the highest point,the football is momentarily at rest before it starts moving downwards. However,the gravitational force continues to act on it. Therefore,its acceleration is equal to the acceleration due to gravity,$g \approx 9.8 \ m/s^2$,directed vertically downwards.
$(b)$ At the highest point,the football momentarily stops before changing its direction of motion. Therefore,its velocity at the highest point is $0 \ m/s$.

(C) Using the principle of conservation of mechanical energy,the potential energy at the top is converted into kinetic energy at the ground.
$PE_{top} = KE_{ground}$
$mgh = \frac{1}{2}mv^2$
$v^2 = 2gh$
Given $g = 10\,ms^{-2}$ and $h = 2\,m$:
$v^2 = 2 \times 10 \times 2 = 40$
$v = \sqrt{40} = 2\sqrt{10}\,ms^{-1}$
Thus,the speed of the body at the ground is $2\sqrt{10}\,ms^{-1}$.

(D) Given: $h = 1\, km = 1000\, m$,$g = 10\, m/s^2$,$\rho = 10^3\, kg/m^3$.
$(a)$ Using $v^2 = u^2 + 2gh$ with $u = 0$: $v = \sqrt{2 \times 10 \times 1000} = \sqrt{20000} \approx 141.4\, m/s$. In $km/h$: $141.4 \times (18/5) \approx 509\, km/h$.
$(b)$ Radius $r = 2\, mm = 2 \times 10^{-3}\, m$. Mass $m = \rho \times (4/3)\pi r^3 = 10^3 \times (4/3) \times 3.14 \times (2 \times 10^{-3})^3 \approx 3.35 \times 10^{-5}\, kg$. Momentum $p = mv = 3.35 \times 10^{-5} \times 141.4 \approx 4.74 \times 10^{-3}\, kg\cdot m/s$.
$(c)$ Time to flatten $t = d/v = (4 \times 10^{-3}) / 141.4 \approx 2.83 \times 10^{-5}\, s$.
$(d)$ Force $F = \Delta p / \Delta t = (4.74 \times 10^{-3}) / (2.83 \times 10^{-5}) \approx 167.5\, N$.
$(e)$ Area of umbrella $A = \pi R^2 = \pi (0.5)^2 \approx 0.785\, m^2$. Separation $s = 5\, cm = 0.05\, m$. Number of drops $N = A / s^2 = 0.785 / (0.05)^2 = 314$. Total force $F_{total} = N \times F \approx 314 \times 167.5 \approx 5.26 \times 10^4\, N$.

(A) Let the velocities of the two balls be $v_1$ and $v_2$ respectively.
Given $v_2 = v$,then $v_1 = 2v$.
Since the vertical gap between the balls remains constant,both balls must be moving with the same velocity at any given time $t > \Delta t$. This implies that the balls have reached their maximum heights and are falling,or they are at the same stage of motion.
However,the condition that the gap remains constant $15\, m$ implies that the difference in their displacements is constant,which occurs when they are both in free fall under gravity after being thrown upwards.
Let $y_1$ and $y_2$ be the displacements from the top of the building: $y_1 = v_1 t - \frac{1}{2} g t^2$ and $y_2 = v_2 (t - \Delta t) - \frac{1}{2} g (t - \Delta t)^2$.
The gap $y_1 - y_2 = 15$. Substituting $v_1 = 2v$ and $v_2 = v$:
$(2v t - \frac{1}{2} g t^2) - (v(t - \Delta t) - \frac{1}{2} g (t - \Delta t)^2) = 15$.
Simplifying,$2vt - \frac{1}{2}gt^2 - vt + v\Delta t + \frac{1}{2}g(t^2 - 2t\Delta t + \Delta t^2) = 15$.
$vt + v\Delta t - g t \Delta t + \frac{1}{2}g \Delta t^2 = 15$.
For the gap to be constant,the coefficient of $t$ must be zero: $v - g \Delta t = 0 \implies v = g \Delta t$.
Substituting $v = 10 \Delta t$ into the constant term: $10 \Delta t^2 + 10 \Delta t^2 - 10 \Delta t^2 + 5 \Delta t^2 = 15 \implies 5 \Delta t^2 = 15 \implies \Delta t^2 = 3 \implies \Delta t = \sqrt{3} \approx 1.732\, s$.
Then $v = 10 \times 1.732 = 17.32\, m/s$. Thus $v_1 = 34.64\, m/s$ and $v_2 = 17.32\, m/s$.