Q.A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.

Showing the 12 most recent of 18 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A body is thrown vertically upwards from the surface of the earth with a velocity of $60\ \mathrm{ms}^{-1}$. The ratio of the displacements of the body during first, second and third seconds of its upward motion is (Acceleration due to gravity $= 10\ \mathrm{ms}^{-2}$) (A) $1 : 3 : 5$ (B) $11 : 9 : 7$ (C) $1 : 1 : 1$ (D) $1 : 2 : 3$

   ›Reveal solutionSolution

   For upward motion under gravity, the displacements in successive seconds are not in the ratio $1:3:5$ (which holds for free fall from rest). Here the body starts upward at $60\ \mathrm{m/s}$, so the distances covered in the 1st, 2nd, and 3rd seconds are $55\ \mathrm{m}$, $45\ \mathrm{m}$, and $35\ \mathrm{m}$, giving the ratio $11:9:7$.

   The classic ratio $1:3:5$ applies only when a body starts from rest and moves under constant acceleration — like free fall downward. But here the body is thrown upward with an initial speed of $60\ \mathrm{m/s}$, and gravity acts downward. So the motion is uniformly decelerated upward. The displacement in each second is not simply $s = \frac12 g t^2$; we must use $s = u t - \frac12 g t^2$ for upward motion, and then find the displacement during each specific second (i.e., the difference in positions at the end and start of that second).

   Let’s work it out.

   1. Displacement in the $n$th second For motion under constant acceleration $a$, the displacement during the $n$th second is

   $s_n = u + \frac{a}{2}(2n-1)$

   where $u$ is initial velocity and $a$ is acceleration (positive if in direction of $u$).

   Here, upward is positive, so $u = +60\ \mathrm{m/s}$ and $a = -g = -10\ \mathrm{m/s}^2$ (since gravity opposes the motion).

   Therefore

   $s_n = 60 + \frac{(-10)}{2}(2n-1) = 60 - 5(2n-1).$

   2. Compute for $n=1,2,3$
      • For $n=1$: $s_1 = 60 - 5(2\cdot1 - 1) = 60 - 5(1) = 55\ \mathrm{m}$
      • For $n=2$: $s_2 = 60 - 5(2\cdot2 - 1) = 60 - 5(3) = 45\ \mathrm{m}$ …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.A body is allowed to fall freely from a height of 120 m from the ground and at the same moment another ball is thrown vertically upwards from the ground such that it reaches a maximum height of 180 m. The time taken for the two bodies to meet is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $5~\mathrm{s}$ (B) $4~\mathrm{s}$ (C) $2~\mathrm{s}$ (D) $3~\mathrm{s}$

   ›Reveal solutionSolution

   The key is to set the positions of the freely falling body and the upward-thrown ball equal as functions of time. Solving the resulting equation gives the meeting time as $4$ seconds.

   Concept & Intuition

   Both bodies move under constant gravity ($g = 10\ \mathrm{m/s^2}$), so their motions are described by the same kinematic equations. The freely falling body starts from rest at 120 m and accelerates downward. The ball thrown upward from the ground has an initial velocity such that it reaches a maximum height of 180 m. At the meeting point, their heights above ground are equal. Instead of solving for the ball’s initial velocity separately, we can use the fact that at its maximum height its velocity is zero — this gives us the initial speed directly.

   Step-by-step solution

   1. Find the initial velocity of the upward-thrown ball For the ball thrown upward, at maximum height $H = 180\ \mathrm{m}$, its velocity is $0$. Using $v^2 = u^2 - 2gH$ (taking upward as positive):

   $0 = u^2 - 2(10)(180) \quad\Rightarrow\quad u^2 = 3600 \quad\Rightarrow\quad u = 60\ \mathrm{m/s}.$

   2. Write position equations for both bodies Let $t$ be the time after release. Take upward as positive and ground as $y=0$.
      • Freely falling body: starts at $y_0 = 120\ \mathrm{m}$, initial velocity $0$, acceleration $-g$:

   $y_1(t) = 120 - \frac{1}{2}gt^2 = 120 - 5t^2.$

   • Upward-thrown ball: starts at $y_0 = 0$, initial velocity $+60\ \mathrm{m/s}$, acceleration $-g$:

   $y_2(t) = 60t - \frac{1}{2}gt^2 = 60t - 5t^2.$

   3. Set positions equal for meeting They meet when $y_1(t) = y_2(t)$:

   $120 - 5t^2 = 60t - 5t^2.$

   The $-5t^2$ terms cancel, leaving:

   $120 = 60t \quad\Rightarrow\quad t = 2\ \mathrm{s}.$

   4. Check the result …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.A stone of mass 10 g is attached to a string of length 80 cm and is rotated in a horizontal circle at height of ‘h’ from the ground with a centripetal acceleration of $125\,\mathrm{ms}^{-2}$. At a moment the string breaks and the stone strikes the ground at a horizontal distance of 10 m. Then the value of ‘h’ is (Acceleration due to gravity $= 10\,\mathrm{ms}^{-2}$) (A) $10\,\mathrm{m}$ (B) $5\,\mathrm{m}$ (C) $15\,\mathrm{m}$ (D) $20\,\mathrm{m}$

   ›Reveal solutionSolution

   The stone’s centripetal acceleration gives its speed; after the string breaks, it becomes a horizontal projectile. Using the horizontal range and speed, we find the time of flight, and from that the height. The value of $h$ is 5 m.

   The problem is a clean combination of circular motion and projectile motion. The stone is moving in a horizontal circle at a constant speed — the centripetal acceleration tells us that speed. When the string breaks, the stone flies off tangentially (horizontally) and falls like a projectile. The horizontal distance it covers before hitting the ground is given, so we can work backwards: from range and speed we get time of flight, and from time of flight we get the height.

   Let’s go step by step.

   1. Find the speed of the stone from centripetal acceleration. Centripetal acceleration $a_c = \frac{v^2}{r}$, where $r$ is the radius of the circle — here the string length $80\,\text{cm} = 0.8\,\text{m}$. Given $a_c = 125\,\text{m/s}^2$,

   $v^2 = a_c \cdot r = 125 \times 0.8 = 100 \quad\Rightarrow\quad v = 10\,\text{m/s}.$

   2. Understand the projectile motion after the string breaks. At the moment of break, the stone has a horizontal velocity $v = 10\,\text{m/s}$ (tangential to the circle) and zero vertical velocity. It falls from height $h$ under gravity $g = 10\,\text{m/s}^2$. The horizontal range $R = 10\,\text{m}$ is given. For a horizontal projectile,

   $R = v \cdot t,$

   where $t$ is the time of flight. So

   $t = \frac{R}{v} = \frac{10}{10} = 1\,\text{s}.$

   3. Relate height to time of flight. For a body dropped from rest (zero initial vertical velocity), the vertical displacement in time $t$ is $h = \frac{1}{2} g t^2.$ …

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4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The angle of inclination of an inclined plane is $45^\circ$. If the coefficient of kinetic friction between the inclined plane and a block on it is $0.2$, then the block released from rest from the top of the inclined plane reaches the bottom of the plane in a time of $3~\text{s}$. If the coefficient of kinetic friction between the block and the inclined plane is $0.8$, then the time taken by the same block to reach the bottom of the plane is (A) $6~\text{s}$ (B) $9~\text{s}$ (C) $7.5~\text{s}$ (D) $4.5~\text{s}$

   ›Reveal solutionSolution

   With equal slide distance, $a_1 t_1^2=a_2 t_2^2$; $t_2=t_1\sqrt{a_1/a_2}=3\sqrt{0.8/0.2}=6\ \text{s}$.

   For a block sliding down a rough incline, the acceleration is

   $a=g(\sin\theta-\mu\cos\theta).$

   At $\theta=45^\circ$, $\sin\theta=\cos\theta=\tfrac{1}{\sqrt2}$, so $a=\dfrac{g}{\sqrt2}(1-\mu)$.

   Case 1 ($\mu_1=0.2$): $a_1=\dfrac{g}{\sqrt2}(0.8)$, reaching the bottom in $t_1=3\ \text{s}$.

   Case 2 ($\mu_2=0.8$): $a_2=\dfrac{g}{\sqrt2}(0.2)$. …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.A ball projected at an angle of $45^\circ$ with the horizontal crosses two points at equal heights separated by a distance at times 2 s and 8 s respectively. The horizontal distance between the two points is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) 300 m (B) 400 m (C) 500 m (D) 600 m

   ›Reveal solutionSolution

   The two equal-height instants are symmetric about the peak, giving $v_x=v_y=50\ \mathrm{m/s}$; the horizontal separation is $v_x(t_2-t_1)=50\times6=300\ \mathrm{m}$.

   Concept

   For a projectile launched at $45^\circ$, the horizontal and vertical launch speeds are equal, and two points at the same height occur at times symmetric about the time of maximum height. Their sum satisfies:

   $t_1 + t_2 = \frac{2v_y}{g}$

   Finding the launch speed

   $2 + 8 = \frac{2v_y}{g} \;\Rightarrow\; 10 = \frac{2v_y}{10} \;\Rightarrow\; v_y = 50\ \mathrm{m/s}$

   Since the angle is $45^\circ$, the horizontal speed equals the vertical speed:

   $v_x = v_y = 50\ \mathrm{m/s}$ …

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6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.At a given place, to increase the number of oscillations made by a simple pendulum in one minute from 72 to 90, the length of the pendulum is to be decreased by (A) 64% (B) 36% (C) 50% (D) 56%

   ›Reveal solutionSolution

   The frequency of a simple pendulum is inversely proportional to the square root of its length. Increasing oscillations per minute from 72 to 90 means frequency increases by a factor of 90/72 = 5/4, so length must decrease by a factor of (4/5)² = 16/25, a reduction of 9/25 = 36%. The correct option is (B).

   Concept & Intuition

   A simple pendulum’s time period $T$ depends only on its length $L$ and gravity $g$:

   $T = 2\pi \sqrt{\frac{L}{g}}$

   The number of oscillations per minute is the frequency $f = \frac{1}{T}$ (in Hz) times 60 seconds, but since we compare counts in the same fixed time, the ratio of counts equals the ratio of frequencies.

   Key insight: frequency is inversely proportional to the square root of length. So if you want more oscillations in the same time, you must shorten the pendulum. The percentage decrease in length follows directly from the ratio of frequencies.

   Step-by-step reasoning

   1. Relate oscillations to frequency Let $n_1 = 72$ and $n_2 = 90$ be the number of oscillations in one minute. Since time is fixed,

   $\frac{n_2}{n_1} = \frac{f_2}{f_1}$

   So

   $\frac{f_2}{f_1} = \frac{90}{72} = \frac{5}{4}$

   2. Express frequency in terms of length For a simple pendulum,

   $f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$

   Hence frequency is proportional to $L^{-1/2}$.

   Therefore,

   $\frac{f_2}{f_1} = \sqrt{\frac{L_1}{L_2}}$

   3. Set up the ratio From step 1 and step 2:

   $\sqrt{\frac{L_1}{L_2}} = \frac{5}{4}$

   Square both sides:

   $\frac{L_1}{L_2} = \left(\frac{5}{4}\right)^2 = \frac{25}{16}$

   So

   $L_2 = \frac{16}{25} L_1$ …

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7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.At a given place, to increase the number of oscillations made by a simple pendulum in one minute from 72 to 90, the length of the pendulum is to be decreased by (A) 50% (B) 36% (C) 64% (D) 56%

   ›Reveal solutionSolution

   The number of oscillations per minute is proportional to the frequency, which is inversely proportional to the square root of the pendulum length. To increase the frequency from 72 to 90 oscillations per minute, the length must be decreased by 36%, so the correct option is (B).

   The key concept here is the simple pendulum’s frequency formula:

   $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$

   where $f$ is the frequency (oscillations per second), $g$ is gravity, and $L$ is the length. Since the number of oscillations in a fixed time (one minute) is directly proportional to frequency, we have $N \propto 1/\sqrt{L}$. This means that if you want more oscillations in the same time, you must shorten the pendulum. The relationship is inverse square root, not linear — a common pitfall is to think a 25% increase in frequency requires a 25% decrease in length, which is wrong.

   Let’s work through it step by step.

   1. Set up the proportionality Let $N$ be the number of oscillations per minute. Then

   $N \propto f \propto \frac{1}{\sqrt{L}}$

   So we can write

   $N = \frac{k}{\sqrt{L}}$

   where $k$ is a constant for the given place (same $g$).

   2. Write the two conditions Initial: $N_1 = 72$, length $L_1$. Final: $N_2 = 90$, length $L_2$. From the proportionality:

   $72 = \frac{k}{\sqrt{L_1}}, \quad 90 = \frac{k}{\sqrt{L_2}}$

   3. Find the ratio of lengths Divide the second equation by the first:

   $\frac{90}{72} = \frac{k/\sqrt{L_2}}{k/\sqrt{L_1}} = \frac{\sqrt{L_1}}{\sqrt{L_2}}$

   Simplify the fraction:

   $\frac{90}{72} = \frac{5}{4}$

   So

   $\frac{\sqrt{L_1}}{\sqrt{L_2}} = \frac{5}{4}$

   Square both sides:

   $\frac{L_1}{L_2} = \left(\frac{5}{4}\right)^2 = \frac{25}{16}$

   Hence

   $L_2 = \frac{16}{25} L_1$ …

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8. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A ball projected at an angle of $45^\circ$ with the horizontal crosses two points at equal heights separated by a distance at times $2\ \mathrm{s}$ and $8\ \mathrm{s}$ respectively. The horizontal distance between the two points is (Acceleration due to gravity $= 10\ \mathrm{ms}^{-2}$) (A) 500 m (B) 600 m (C) 400 m (D) 300 m

   ›Reveal solutionSolution

   The key idea is that for a projectile launched at $45^\circ$, the times to reach the same height are symmetric about the time of maximum height. Using the given times $2$ s and $8$ s, we find the time of flight and horizontal velocity, then compute the horizontal distance between the two points as $400$ m.

   Concept and Intuition

   When a projectile is launched at $45^\circ$, its trajectory is symmetric. If two points are at the same height, the times at which the projectile passes them are equally spaced around the time of maximum height. This symmetry lets us find the total time of flight and the horizontal velocity from the given times, without needing the initial speed explicitly.

   Step-by-step solution

   1. Understand the symmetry For a projectile, the vertical motion is symmetric: the time to go up to a certain height equals the time to come back down to that same height. If the projectile passes a given height at times $t_1$ and $t_2$ (with $t_2 > t_1$), then the time of maximum height is exactly halfway:

   $t_{\text{max}} = \frac{t_1 + t_2}{2}.$

   Here $t_1 = 2$ s and $t_2 = 8$ s, so

   $t_{\text{max}} = \frac{2 + 8}{2} = 5\ \text{s}.$

   2. Find the total time of flight The total time of flight $T$ is twice the time to reach maximum height (since the projectile goes up and then down):

   $T = 2 \times t_{\text{max}} = 2 \times 5 = 10\ \text{s}.$

   3. Determine the horizontal velocity The horizontal component of velocity is constant. For a launch at $45^\circ$, the initial speed $u$ satisfies $u_x = u \cos 45^\circ = \frac{u}{\sqrt{2}}$. The range $R$ (total horizontal distance) is

   $R = u_x \cdot T.$

   But we also know the range formula for a projectile:

   $R = \frac{u^2 \sin 2\theta}{g}.$

   With $\theta = 45^\circ$, $\sin 90^\circ = 1$, so

   $R = \frac{u^2}{g}.$

   Equating the two expressions for $R$:

   $u_x \cdot T = \frac{u^2}{g}.$

   Since $u_x = \frac{u}{\sqrt{2}}$ and $T = 10$ s, $g = 10$ m/s², we get

   $\frac{u}{\sqrt{2}} \cdot 10 = \frac{u^2}{10}.$

   Cancel $u$ (assuming $u \neq 0$):

   $\frac{10}{\sqrt{2}} = \frac{u}{10} \quad \Rightarrow \quad u = \frac{100}{\sqrt{2}} = 50\sqrt{2}\ \text{m/s}.$

   Then

   $$$$

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9. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The vertical displacement (y in metre) of a projectile in terms of its horizontal displacement (x in metre) is given by $y = (\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity $= 10 \ \mathrm{ms}^{-2}$) (A) $5\sqrt{3} \ \mathrm{s}$ (B) $\sqrt{3} \ \mathrm{s}$ (C) $0.2 \ \mathrm{s}$ (D) $0.2\sqrt{3} \ \mathrm{s}$

   ›Reveal solutionSolution

   The trajectory equation $y = \sqrt{3}\,x - 0.2\,x^2$ is of the form $y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}\,x^2$. Comparing coefficients gives $\tan\theta = \sqrt{3}$ and $\frac{g}{2u^2\cos^2\theta} = 0.2$. Solving yields $u = 10$ m/s and $\theta = 60^\circ$, so time of flight $T = \frac{2u\sin\theta}{g} = \sqrt{3}$ s. The correct option is (B).

   ---

   The key idea is that the given equation is the trajectory equation of a projectile. In standard form, it relates horizontal and vertical displacements directly, and its coefficients encode the launch speed and angle. By matching coefficients, we can extract the initial velocity and angle, then compute the time of flight.

   Why this works:

   The trajectory equation is derived by eliminating time from the parametric equations $x = u\cos\theta \cdot t$ and $y = u\sin\theta \cdot t - \frac12 g t^2$. The result is a quadratic in $x$:

   $y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}\,x^2.$

   So any quadratic $y = ax - bx^2$ (with no constant term) that describes projectile motion must have $a = \tan\theta$ and $b = \frac{g}{2u^2\cos^2\theta}$. We simply match numbers.

   ---

   Step-by-step solution:

   1. Identify the standard form The general trajectory equation is

   $y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}\,x^2.$

   Here $g = 10$ m/s². Our given equation is

   $y = \sqrt{3}\,x - 0.2\,x^2.$

   2. Match the coefficient of $x$ The coefficient of $x$ is $\tan\theta$. So

   $\tan\theta = \sqrt{3} \quad\Rightarrow\quad \theta = 60^\circ.$

   3. Match the coefficient of $x^2$ The coefficient of $x^2$ is $-\frac{g}{2u^2\cos^2\theta}$. Thus

   $\frac{g}{2u^2\cos^2\theta} = 0.2.$

   Substitute $g = 10$ and $\cos 60^\circ = \frac12$:

   $\frac{10}{2u^2 \cdot \left(\frac12\right)^2} = 0.2.$

   Simplify: $\left(\frac12\right)^2 = \frac14$, so denominator is $2u^2 \cdot \frac14 = \frac{u^2}{2}$. Hence …

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10. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
    Q.A solid sphere of radius 18 cm is rolling down from rest from the top of an inclined plane of length 14 m and angle of inclination $30^\circ$. The time taken by the sphere to reach the bottom of the inclined plane is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) 2.8 s (B) 4.2 s (C) 3.5 s (D) 1.4 s

    ›Reveal solutionSolution

    For a rolling rigid body, the acceleration down an incline is $a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$. For a solid sphere, $I = \frac{2}{5}mR^2$, so $a = \frac{5}{7}g\sin\theta$. Using $s = \frac12 a t^2$ with $s = 14$ m, $\theta = 30^\circ$, $g = 10$ m/s² gives $t = 2.8$ s. The correct option is (A).

    The key insight is that when a sphere rolls without slipping, its kinetic energy is split between translation and rotation. The rotational inertia resists the spin-up, so the sphere accelerates slower than a block sliding frictionlessly. The moment of inertia for a solid sphere is $\frac{2}{5}mR^2$, and this factor directly reduces the effective acceleration.

    Let’s work through it step by step.

    1. Identify the acceleration of a rolling body on an incline. For pure rolling, the torque about the centre is provided by friction, and the linear acceleration $a$ satisfies

    $a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}.$

    This comes from combining Newton’s second law for translation ($mg\sin\theta - f = ma$) and rotation ($fR = I\alpha$, with $\alpha = a/R$). The friction $f$ cancels out, leaving only the geometry and inertia.

    2. Plug in the moment of inertia for a solid sphere. For a solid sphere, $I = \frac{2}{5}mR^2$. Then

    $1 + \frac{I}{mR^2} = 1 + \frac{2}{5} = \frac{7}{5}.$

    So the acceleration becomes

    $a = \frac{g\sin\theta}{7/5} = \frac{5}{7}g\sin\theta.$

    3. Substitute the given values. $\theta = 30^\circ$, so $\sin 30^\circ = \frac12$. And $g = 10$ m/s².

    $a = \frac{5}{7} \times 10 \times \frac12 = \frac{5}{7} \times 5 = \frac{25}{7}~\mathrm{m/s^2}.$

    4. Use the kinematic equation for constant acceleration from rest. …

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11. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.Two bodies A and B are projected simultaneously with velocities $20~\mathrm{ms}^{-1}$ and $40~\mathrm{ms}^{-1}$ respectively. Body A is projected vertically up from the top of a tower of height $80~\mathrm{m}$ and body B is projected vertically up from the bottom of the same tower. The bodies A and B meet in time of (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $5~\mathrm{s}$ (B) $3~\mathrm{s}$ (C) $6~\mathrm{s}$ (D) $4~\mathrm{s}$

    ›Reveal solutionSolution

    The key idea is to set up the position equations for both bodies from a common reference (the ground) and equate them at the meeting point. The time of meeting is 4 seconds.

    The problem involves two objects moving under uniform acceleration (gravity). Both are projected vertically upward, but from different heights — one from the top of an 80 m tower, the other from the ground. They meet somewhere in between. The natural approach is to write their positions as functions of time from a single reference point, then solve for the time when those positions are equal.

    Why does this work? Because under constant acceleration, the position of any object is given by $s = s_0 + u t + \frac{1}{2} a t^2$. Here, acceleration is $-g$ (taking upward as positive). If we measure everything from the ground, the initial positions are different, but the same physics applies to both.

    1. Choose a reference and define variables.

       Let upward be the positive direction. Take the ground as the origin ($y = 0$).

       For body A (from the tower top): initial height $y_{0A} = 80$ m, initial velocity $u_A = +20$ m/s, acceleration $a = -g = -10$ m/s².

       For body B (from the ground): initial height $y_{0B} = 0$ m, initial velocity $u_B = +40$ m/s, same acceleration.

    2. Write the position equations.

       Using $y(t) = y_0 + u t + \frac{1}{2} a t^2$:

    $y_A(t) = 80 + 20t - \frac{1}{2}(10)t^2 = 80 + 20t - 5t^2$

    $y_B(t) = 0 + 40t - \frac{1}{2}(10)t^2 = 40t - 5t^2$

    3. Set the positions equal at the meeting time. They meet when $y_A(t) = y_B(t)$:

    $80 + 20t - 5t^2 = 40t - 5t^2$

    4. Simplify and solve. The $-5t^2$ terms cancel on both sides, leaving:

    $80 + 20t = 40t$

    $80 = 20t$ …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.A stone is thrown vertically up from the top end of a window of height $1.8\,\mathrm{m}$ with a velocity of $8\,\mathrm{ms}^{-1}$. The time taken by the stone to cross the window during its downward journey is (Acceleration due to gravity $= 10\,\mathrm{ms}^{-2}$) (A) $0.8\,\mathrm{s}$ (B) $1.6\,\mathrm{s}$ (C) $1.0\,\mathrm{s}$ (D) $0.2\,\mathrm{s}$

    ›Reveal solutionSolution

    The stone crosses the window twice — once going up and once coming down. The time to cross it on the way down is found by solving for the two instants when the stone is at the top and bottom of the window, then taking the difference. The answer is 0.2 s.

    ---

    Concept and Intuition

    The stone is thrown upward from the top of the window. It will rise, stop, then fall back down. On the way down, it will pass the window again — first the top edge, then the bottom edge. The question asks for the time it takes to go from the top edge to the bottom edge during the downward journey.

    We can model the motion using the equation for displacement under constant acceleration. The key trick: treat the top of the window as the origin ($y = 0$), upward as positive. Then the bottom of the window is at $y = -1.8\,\mathrm{m}$. The stone starts at $y = 0$ with initial velocity $u = +8\,\mathrm{m/s}$ and acceleration $a = -10\,\mathrm{m/s}^2$.

    We find the times when the stone is at $y = 0$ (top) and $y = -1.8$ (bottom) on the way down, and subtract.

    ---

    Step-by-step solution

    1. Set up the equation of motion The displacement $s$ from the starting point after time $t$ is:

    $s = ut + \frac{1}{2} a t^2$

    With $u = 8$, $a = -10$:

    $s = 8t - 5t^2$

    2. Find the times when the stone is at the top of the window ($s = 0$)

    $0 = 8t - 5t^2 \quad \Rightarrow \quad t(8 - 5t) = 0$

    So $t = 0$ (start) or $t = \frac{8}{5} = 1.6\,\mathrm{s}$.

    The stone returns to the top of the window at $t = 1.6\,\mathrm{s}$ on its way down.

    3. Find the times when the stone is at the bottom of the window ($s = -1.8$)

    $-1.8 = 8t - 5t^2$

    Rearranging:

    $5t^2 - 8t - 1.8 = 0$

    Multiply by 5 to avoid decimals:

    $25t^2 - 40t - 9 = 0$

    Solve using the quadratic formula:

    $t = \frac{40 \pm \sqrt{1600 + 900}}{50} = \frac{40 \pm \sqrt{2500}}{50} = \frac{40 \pm 50}{50}$

    So:

    $t = \frac{90}{50} = 1.8\,\mathrm{s} \quad \text{or} \quad t = \frac{-10}{50} = -0.2\,\mathrm{s}$ …

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