Q.A gun can fire shells with a maximum speed $v_0$. On level ground the greatest horizontal range it can achieve (firing at $45^\circ$) is $R = \dfrac{v_0^2}{g}$. A target lies farther away, a distance $\Delta x$ beyond $R$ — that is, at a horizontal distance $R + \Delta x$ from the gun. Show that this target can still be hit with the same gun by raising the gun to a height of at least

[!FORMULA] $h = \Delta x\left[1 + \frac{\Delta x}{R}\right].$

(Approach: place the gun at the top of a tower of height $h$; take the launch point as the origin with $x$ horizontal and $y$ vertically upward. The target then sits at $x = R + \Delta x$ and $y = -h$. Fire at the angle that maximises horizontal reach, $45^\circ$.)

Showing the 12 most recent of 20 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A boy, standing at a certain height, kicks a football horizontally with a velocity of $19.6 \mathrm{~ms}^{-1}$. What will be the ratio of horizontal and vertical components of velocities after $2 s$ ? (A) $1: \sqrt{2}$ (B) $\left(\frac{\sqrt{3}}{2}\right): 1$ (C) $1: 1$ (D) $1:0.5$

   ›Reveal solutionSolution

   The horizontal velocity remains constant, while the vertical velocity increases due to gravity. After 2 seconds, both components are equal, so the ratio is 1:1.

   Concept & Intuition

   When a football is kicked horizontally from a height, it has an initial vertical velocity of zero. Gravity acts only downward, so the horizontal component stays unchanged (no horizontal force, ignoring air resistance). The vertical component grows linearly with time as $v_y = g t$. The question asks for the ratio after exactly 2 seconds — a direct calculation will show they match.

   Step-by-step solution

   1. Identify the initial velocities

      The ball is kicked horizontally at $19.6 \, \text{m/s}$.

      • Horizontal component: $u_x = 19.6 \, \text{m/s}$ (constant).
      • Vertical component: $u_y = 0 \, \text{m/s}$ (starts from rest vertically).

   2. Horizontal velocity after 2 seconds

      No horizontal acceleration:

   $v_x = u_x = 19.6 \, \text{m/s}$

   3. Vertical velocity after 2 seconds Using $v_y = u_y + g t$ with $g = 9.8 \, \text{m/s}^2$ (downward positive):

   $v_y = 0 + (9.8)(2) = 19.6 \, \text{m/s}$

   4. Compute the ratio …

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2. KCET 2026Set C21 markMCQ
   Q.Two bodies are projected with the same velocity. If one is projected at an angle of $30^\circ$ and the other at $45^\circ$ to the horizontal, then the ratio of maximum heights attained is (A) 3:1 (B) 1:2 (C) 4:1 (D) 1:3

   ›Reveal solutionSolution

   The maximum height in projectile motion is $H=\dfrac{u^2\sin^2\theta}{2g}$; for equal launch speeds, the ratio of maximum heights equals the ratio of $\sin^2\theta$ for the two angles.

   Step 1 — Write the maximum-height formula

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

   Both bodies are projected with the same speed $u$ and experience the same $g$, so these cancel in a ratio.

   Step 2 — Compute the ratio …

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3. KCET 2025Set D-41 markMCQ
   Q.In the projectile motion of a particle on a level ground, which of the following remains constant with reference to time and position? (A) Average velocity between any two points on the path (B) Horizontal component of velocity (C) Angle between the instantaneous velocity with the horizontal (D) Vertical component of the velocity of the projectile

   ›Reveal solutionSolution

   Gravity has no horizontal component, so $a_x = 0$ and $v_x$ is constant throughout the flight; everything else in the options changes.

   Step 1 — Resolve the motion.

   Launch a projectile with speed $u$ at angle $\theta$ to the horizontal. The only force acting (neglecting air resistance) is gravity, which points straight down. Resolving:

   $a_x = 0, \qquad a_y = -g$

   This single fact — that gravity is purely vertical — is the whole of the physics here. Horizontal and vertical motions are independent.

   Step 2 — Integrate each component.

   Horizontal: with $a_x = 0$,

   $v_x = u\cos\theta = \text{constant}$

   It does not depend on $t$, nor on where the particle is. This is the invariant.

   Vertical: with $a_y = -g$,

   $v_y = u\sin\theta - g t$

   This changes continuously — positive on the way up, zero at the apex, negative on the way down.

   Step 3 — Test each option.

   1. (A) Average velocity between any two points on the path.

   $\vec{v}_{\text{avg}} = \frac{\Delta \vec{r}}{\Delta t}$

   Its horizontal part is indeed always $u\cos\theta$, but its vertical part is $\tfrac{\Delta y}{\Delta t}$, which differs for different pairs of points (e.g. rising portion vs. falling portion). So the average velocity (a vector) is not constant. ✗

   2. (B) Horizontal component of velocity. $v_x = u\cos\theta$ for all $t$ and at every position — constant. ✓ …

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.A boy is running along a straight horizontal road with a constant speed $5 \mathrm{~ms}^{-1}$. While running he throws a stone with a velocity $30 \mathrm{~ms}^{-1}$ at an angle $60^{\circ}$ with the horizontal. Then the time of flight of the stone is: (Given $g=10 \mathrm{~ms}^{-2}$ ) (A) $3 \sqrt{2}$ (B) $\frac{\sqrt{3}}{2}$ (C) $4 \sqrt{3}$ (D) $3 \sqrt{3}$

   ›Reveal solutionSolution

   The stone’s time of flight depends only on the vertical component of its absolute velocity (relative to ground). The boy’s horizontal motion adds to the stone’s horizontal velocity but does not affect the vertical launch speed. The vertical component is $30 \sin 60^\circ = 15\sqrt{3} \, \text{m/s}$, so time of flight $T = \frac{2u_y}{g} = 3\sqrt{3} \, \text{s}$, matching option (D).

   ---

   Concept & Intuition

   The key idea: time of flight for a projectile depends only on the vertical component of its initial velocity relative to the ground, and on gravity. The boy is running horizontally, so his velocity adds to the stone’s horizontal component but does not change the vertical component the stone is given at the moment of release. Therefore, we can ignore the boy’s motion when computing the time the stone stays in the air.

   ---

   Step-by-step solution

   1. Identify the relevant velocity components The stone is thrown with speed $30 \, \text{m/s}$ at $60^\circ$ to the horizontal, relative to the boy. But the boy himself is moving horizontally at $5 \, \text{m/s}$. The stone’s absolute velocity (relative to ground) is the vector sum:

   $\vec{v}_{\text{stone}} = \vec{v}_{\text{boy}} + \vec{v}_{\text{throw}}$

   The boy’s velocity is purely horizontal: $\vec{v}_{\text{boy}} = (5, 0)$.

   2. Find the vertical component The throw gives the stone a vertical component:

   $v_y = 30 \sin 60^\circ = 30 \cdot \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{m/s}$

   The boy’s motion adds nothing vertically, so this is the stone’s absolute vertical launch speed.

   3. Apply the time-of-flight formula For a projectile launched from ground level (or returning to the same height), the time of flight is: $T = \frac{2 v_y}{g}$ …

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5. COMEDK 2025Set 2025-M1 markMCQ
   Q.Three identical conducting balls $\mathrm{A}, \mathrm{B}$ and C , each of mass m , are thrown upward at an angle $\Theta$ to the horizontal with an initial speed $v$ in a region of space that has a uniform electric field $E$ downward along with the gravitational field $g . A$ is positively charged, B is uncharged and C is negatively charged. Rank the ranges R of these three balls in increasing order. (A) $R_A< R_B< R_C$ (B) $R_B< R_C< R_A$ (C) $R_A=R_B< R_C$ (D) $R_C< R_B< R_A$

   ›Reveal solutionSolution

   The key idea is that the net vertical acceleration differs for each ball due to the electric field, while the horizontal motion is unaffected; the ball with the smallest downward acceleration stays in the air longest and thus has the greatest range. The correct order is $R_C < R_B < R_A$, so option (D).

   We have three balls: A (positive charge), B (uncharged), and C (negative charge). They are all launched identically — same speed $v$, same angle $\Theta$ to the horizontal — in a region with both gravity $g$ downward and a uniform electric field $E$ also downward. The only difference is their charge, which affects the net vertical force.

   Concept and intuition:

   The range of a projectile depends on its time of flight and horizontal velocity. Since the horizontal velocity is the same for all (no horizontal force), the range is proportional to the time of flight. The time of flight is determined by the vertical motion: how long it takes to go up and come back down. A larger downward acceleration means the ball returns sooner, reducing range. Here, the electric field adds an extra downward force on positive charges, reduces downward force on negative charges (since it’s opposite), and leaves uncharged balls unaffected. So A (positive) has the largest downward acceleration, C (negative) the smallest, and B is in between.

   Let’s work through it step by step.

   1. Identify the forces and net vertical acceleration for each ball.
      • For ball A (positive charge $q$): The electric force is $qE$ downward, same direction as gravity. So net downward force = $mg + qE$. Acceleration downward:

   $a_A = g + \frac{qE}{m}$

   • For ball B (uncharged): No electric force. Acceleration downward:

   $a_B = g$

   • For ball C (negative charge $-q$): The electric force is $qE$ upward (opposite to the field direction). So net downward force = $mg - qE$. Acceleration downward:

   $a_C = g - \frac{qE}{m}$

   (We assume $mg > qE$ so the ball still goes upward initially; otherwise it wouldn’t rise.)

   2. Relate time of flight to vertical acceleration. The initial vertical velocity is $v_y = v \sin\Theta$. The time to reach maximum height is $t_{\text{up}} = v_y / a$, and the total time of flight is twice that (since symmetric return under constant acceleration):

   $T = \frac{2 v \sin\Theta}{a}$

   So a larger $a$ gives a shorter time of flight.

   3. Compare the accelerations. Clearly:

   $a_A > a_B > a_C$

   Therefore:

   $T_A < T_B < T_C$

   4. Range depends on time of flight. …

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6. KCET 2024Set D-21 markMCQ
   Q.An athlete runs along a circular track of diameter 80 m. The distance travelled and the magnitude of displacement of the athlete when he covers $\dfrac{3^{\text{th}}}{4}$ of the circle is (in m) (A) $60\pi, 40\sqrt{2}$ (B) $40\pi, 60\sqrt{2}$ (C) $120\pi, 80\sqrt{2}$ (D) $80\pi, 120\sqrt{2}$

   ›Reveal solutionSolution

   The athlete runs three-quarters of a circle, so distance is $\frac{3}{4}$ of the circumference, and displacement is the chord connecting start to end — a right triangle with legs equal to the radius. The answer is (A) $60\pi, 40\sqrt{2}$.

   The key idea here is the difference between distance (a scalar: total path length) and displacement (a vector: straight-line separation from start to finish). For circular motion, distance is a fraction of the circumference; displacement depends on the chord length between the initial and final positions.

   When the athlete covers $\frac{3}{4}$ of the circle, he starts at some point and goes three-quarters of the way around. That leaves him exactly one-quarter of the circle away from the start — a $90^\circ$ angular separation. The chord across a $90^\circ$ arc is the hypotenuse of an isosceles right triangle whose equal sides are the radius.

   Let’s work it out step by step.

   1. Find the radius.

      Diameter $= 80$ m, so radius $r = \frac{80}{2} = 40$ m.

   2. Distance travelled.

      Circumference of the full circle $= 2\pi r = 2\pi \times 40 = 80\pi$ m.

      For $\frac{3}{4}$ of the circle, distance $= \frac{3}{4} \times 80\pi = 60\pi$ m.

   3. Displacement magnitude. …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.Two balls are thrown horizontally, one from the window of first floor which is $3 \mathrm{~m}$ high from the ground and second from the second floor which is $6 \mathrm{~m}$ high from the ground, of a multi storey building, with the same speed of $6 \mathrm{~ms}^{-1}$. Calculate the distance that will separate the two balls when they hit the ground. (A) 2.28 m (B) 1.94 m (C) 1.81 m (D) 0.39 m

   ›Reveal solutionSolution

   Both balls are thrown horizontally from different heights with the same speed; their horizontal separation when they land is simply the difference in their horizontal ranges, which depends only on the difference in their times of flight. The answer is about 1.81 m.

   Concept & Intuition

   When an object is thrown horizontally from a height, its motion is a combination of constant horizontal velocity (no horizontal acceleration) and free fall vertically (constant downward acceleration $g$). The time to hit the ground depends only on the vertical drop, not on the horizontal speed. Since both balls have the same horizontal speed, the one that falls longer will travel farther horizontally. The separation between them when they land is just the difference in their horizontal distances traveled.

   ---

   1. Find the time of flight for each ball For a ball dropped from height $h$ with zero initial vertical velocity, the vertical motion satisfies

   $h = \frac{1}{2} g t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2h}{g}}$

   Take $g = 9.8 \, \text{m/s}^2$.

   • First floor: $h_1 = 3 \, \text{m}$

   $t_1 = \sqrt{\frac{2 \times 3}{9.8}} = \sqrt{\frac{6}{9.8}} \approx \sqrt{0.6122} \approx 0.7825 \, \text{s}$

   • Second floor: $h_2 = 6 \, \text{m}$

   $t_2 = \sqrt{\frac{2 \times 6}{9.8}} = \sqrt{\frac{12}{9.8}} \approx \sqrt{1.2245} \approx 1.1066 \, \text{s}$

   2. Compute the horizontal range for each ball

      Horizontal velocity is constant: $v_x = 6 \, \text{m/s}$.

      Range $R = v_x \times t$.

      • Ball from first floor:

   $R_1 = 6 \times 0.7825 \approx 4.695 \, \text{m}$

   • Ball from second floor:

   $R_2 = 6 \times 1.1066 \approx 6.6396 \, \text{m}$

   3. Find the separation distance The balls land at different horizontal distances from the building. The separation is the difference: $$$$

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.A cricketer of height $2.5 \mathrm{~m}$ throws a ball at an angle of $30^{\circ}$ with the horizontal such that it is received by another cricketer of same height standing at a distance of $50 \mathrm{~m}$ from the first one. The maximum height attained by the ball is ($\tan 30^{\circ}=0.577$) (A) 7.9 m (B) 10 m (C) 10.7 m (D) 9.7 m

   ›Reveal solutionSolution

   Launch and catch heights are equal, so the range formula gives max height above the throw point $H = \tfrac{R\tan\theta}{4}=7.21\text{ m}$; adding the launch height $2.5$ m gives $\approx 9.7$ m above the ground.

   Both cricketers have the same height ($2.5$ m), so the ball leaves and is caught at the same level. The horizontal separation is the range $R = 50$ m at angle $\theta = 30^\circ$.

   For a projectile that returns to its launch level, the maximum height above the launch point relates to the range by

   $H = \frac{R\tan\theta}{4}.$

   (Derivation: $R = \dfrac{u^2\sin 2\theta}{g}$ and $H = \dfrac{u^2\sin^2\theta}{2g}$, so $\dfrac{H}{R} = \dfrac{\sin^2\theta}{2\sin 2\theta} = \dfrac{\tan\theta}{4}$.) …

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.A metal ball of $20 \mathrm{~g}$ is projected at an angle $30^{\circ}$ with the horizontal with an initial velocity $10 \mathrm{~ms}^{-1}$. If the mass and angle of projection are doubled keeping the initial velocity the same, the ratio of the maximum height attained in the former to the latter case is : (A) 1 : 2 (B) 2 : 1 (C) 1 : 3 (D) 3 : 1

   ›Reveal solutionSolution

   The maximum height in projectile motion depends on the square of the vertical component of velocity. Doubling both mass and angle changes only the angle, so the ratio of heights is determined by $\sin^2\theta$ — giving $1:3$.

   Concept & Intuition

   The maximum height $H$ of a projectile depends only on the vertical component of the initial velocity: $H = \frac{u^2 \sin^2\theta}{2g}$. Mass does not appear — it cancels out in the energy or kinematic derivation. So when the problem says “mass and angle are doubled”, the mass change is irrelevant. Only the angle change matters: from $30^\circ$ to $60^\circ$. The ratio of heights is therefore the ratio of $\sin^2(30^\circ)$ to $\sin^2(60^\circ)$.

   Step-by-step reasoning

   1. Recall the formula for maximum height For a projectile launched with speed $u$ at angle $\theta$ above horizontal, the maximum height is

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

   This comes from setting the vertical velocity to zero at the peak: $0 = (u\sin\theta)^2 - 2gH$.

   2. Identify the two cases

      • Former case: mass $m_1 = 20\ \text{g}$, angle $\theta_1 = 30^\circ$, speed $u = 10\ \text{m/s}$.
      • Latter case: mass $m_2 = 40\ \text{g}$ (doubled), angle $\theta_2 = 60^\circ$ (doubled), same speed $u = 10\ \text{m/s}$.

   3. Notice that mass does not appear in $H$

      The formula for $H$ contains only $u$, $\theta$, and $g$. So the change in mass has zero effect on the height. This is a common trick — students might think heavier objects rise less, but in projectile motion (ignoring air resistance), all objects follow the same parabolic path for the same initial velocity.

   4. Write the heights …

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10. KCET 2023Set A-31 markMCQ
    Q.A ball of mass $0.2\, \mathrm{kg}$ is thrown vertically down from a height of $10\, \mathrm{m}$. It collides with the floor and loses $50\%$ of its energy and then rises back to the same height. The value of its initial velocity is (A) $14\, \mathrm{ms}^{-1}$ (B) $196\, \mathrm{ms}^{-1}$ (C) $20\, \mathrm{ms}^{-1}$ (D) zero

    ›Reveal solutionSolution

    Write the energy just before impact, halve it, and set it equal to the potential energy needed to climb back to $10$ m.

    Step 1 — Energy just before hitting the floor.

    Taking the floor as the reference level and letting $u$ be the (downward) initial speed at height $h=10$ m, conservation of mechanical energy during the fall gives

    $E_{\text{before}} = \frac{1}{2}mu^{2} + mgh.$

    Step 2 — The collision.

    The ball loses $50\%$ of its energy, so it leaves the floor with

    $E_{\text{after}} = \frac{1}{2}\left(\frac{1}{2}mu^{2} + mgh\right).$

    Step 3 — Condition for rising back to the same height.

    To just reach height $h$ again (arriving with zero speed), all of $E_{\text{after}}$ must convert to potential energy:

    $\frac{1}{2}\left(\frac{1}{2}mu^{2} + mgh\right) = mgh.$ …

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11. COMEDK 2023Set 2023-E1 markMCQ
    Q.A hockey player hits the ball at an angle of $37^{\circ}$ from the horizontal with an initial speed of $40 \mathrm{~m} / \mathrm{s}$ (a right angled triangle with one of the angle is $37^{\circ}$ and their sides in the ratio of $6: 8: 10$). Assume that the ball is in a vertical plane. The time at which the ball reaches the highest point of its path is (A) 2.4 s (B) 0.32 s (C) 3.2 s (D) 0.24 s

    ›Reveal solutionSolution

    At the top the vertical velocity is zero, so $t=u\sin\theta/g=40(0.6)/10=2.4\ \text{s}$.

    For projectile motion, the ball reaches its highest point when the vertical component of velocity becomes zero:

    $t = \frac{u\sin\theta}{g}$ …

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12. COMEDK 2023Set 2023-M1 markMCQ
    Q.A particle is projected at an angle $30^{\circ}$ with horizontal having kinetic energy $K$. The kinetic energy of the particle at highest point is. (A) $\frac{1}{2} K$ (B) $\frac{3}{4} K$ (C) $\frac{3}{8} K$ (D) $\frac{5}{8} K$

    ›Reveal solutionSolution

    At the top of the trajectory the vertical velocity is zero; only $v\cos30^\circ$ remains, so the KE falls to $K\cos^2 30^\circ=\tfrac34 K$.

    Initial KE: $K=\tfrac12 mv^2$.

    At the highest point the vertical component of velocity is zero; the horizontal component $v\cos30^\circ$ is unchanged. So: …

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