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 69 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the minimum velocity of a projectile during its motion is $40\ ms^{-1}$ and the ratio of its vertical and horizontal displacements at a time of 2s is 1:2, then the angle of projection of the projectile is (Acceleration due to gravity $= 10\ ms^{-2}$) (A) $\sin^{-1}(0.6)$ (B) $\cos^{-1}(0.6)$ (C) $\tan^{-1}(0.6)$ (D) $\sec^{-1}(0.6)$

   ›Reveal solutionSolution

   The minimum projectile speed equals the horizontal velocity component; combining that with the given displacement ratio at $t=2$s gives a 3-4-5 velocity triangle, so $\theta = \sin^{-1}(0.6)$.

   Concept and Intuition

   During projectile motion, the horizontal velocity component ($u\cos\theta$) stays constant throughout the flight, while the vertical component continuously changes (and is zero at the peak). The speed is minimum exactly at the peak, where the velocity is purely horizontal — so the minimum speed directly gives us $u\cos\theta$. Combined with a displacement-ratio condition at a specific time, we get two equations for $u\sin\theta$ and $u\cos\theta$, pinning down the angle via a right-triangle ratio.

   Step-by-Step Solution

   1. Minimum velocity during projectile motion = horizontal component (constant throughout flight) = $u\cos\theta = 40\ ms^{-1}$.
   2. At $t=2$s, horizontal displacement: $x = (u\cos\theta)\,t = 40\times2 = 80$ m.
   3. Given the ratio of vertical to horizontal displacement at $t=2$s is $1:2$, so $y = \dfrac{x}{2} = 40$ m.
   4. Vertical displacement formula: $y = u\sin\theta\cdot t - \dfrac12 g t^2$. Substituting $t=2$, $g=10$:

   $40 = 2\,u\sin\theta - \frac12(10)(4) = 2u\sin\theta - 20$

   5. Solve: $2u\sin\theta = 60 \Rightarrow u\sin\theta = 30$.
   6. Now we have $u\cos\theta=40$ and $u\sin\theta=30$. So: $u = \sqrt{(u\cos\theta)^2+(u\sin\theta)^2} = \sqrt{40^2+30^2} = \sqrt{1600+900}=\sqrt{2500}=50\ ms^{-1}$ …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A body is projected with a velocity of $15\sqrt{3}\ ms^{-1}$ at an angle of $60^\circ$ with the horizontal and another body is projected simultaneously from the same point in the same vertical plane with a velocity of $40\ ms^{-1}$ at an angle of $30^\circ$ with the horizontal. The time at which the velocity vector of the two bodies will be in the same direction is (Acceleration due to gravity $= 10\ ms^{-2}$) (A) 3.2 s (B) 2.4 s (C) 1.2 s (D) 3.6 s

   ›Reveal solutionSolution

   Setting the velocity-vector slopes of the two projectiles equal (same direction condition) and solving for $t$ gives $t=2.4$ s.

   Concept and Intuition

   Two velocity vectors point in the same direction exactly when they are parallel, i.e. when their vertical-to-horizontal component ratios (slopes) are equal. Since each projectile's horizontal velocity is constant and its vertical velocity decreases linearly with time due to gravity, equating the two slopes as functions of $t$ gives a single linear equation to solve.

   Step-by-Step Solution

   1. Body A: $u_A=15\sqrt3\ ms^{-1}$ at $\theta_A=60°$.
      • $v_{xA} = u_A\cos60° = 15\sqrt3\times\dfrac12 = 7.5\sqrt3$ (constant)
      • $v_{yA}(t) = u_A\sin60° - gt = 15\sqrt3\times\dfrac{\sqrt3}{2}-10t = \dfrac{15\times3}{2}-10t = 22.5-10t$
   2. Body B: $u_B=40\ ms^{-1}$ at $\theta_B=30°$.
      • $v_{xB} = u_B\cos30° = 40\times\dfrac{\sqrt3}{2}=20\sqrt3$ (constant)
      • $v_{yB}(t) = u_B\sin30°-gt = 40\times\dfrac12-10t = 20-10t$
   3. The velocity vectors are parallel (same direction) when the slopes match:

   $\frac{v_{yA}}{v_{xA}} = \frac{v_{yB}}{v_{xB}} \;\Longrightarrow\; \frac{22.5-10t}{7.5\sqrt3} = \frac{20-10t}{20\sqrt3}$

   4. The $\sqrt3$ cancels from both sides: $\frac{22.5-10t}{7.5} = \frac{20-10t}{20}$ …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.A ball is projected upwards from the top of a tower with a velocity $50\ \mathrm{ms^{-1}}$ making an angle $30°$ with the horizontal. The height of tower is 70 m. After how many seconds from the instant of throwing, will the ball reach the ground? ($g = 10\ \mathrm{ms^{-2}}$) (A) 7 s (B) 9 s (C) 5 s (D) 2 s

   ›Reveal solutionSolution

   Only the vertical component of the initial velocity matters for the time to reach the ground; solving the standard $s=ut-\tfrac12gt^2$ quadratic gives $t=7$ s — option (A).

   Concept and Intuition

   Projectile motion problems separate into independent horizontal and vertical components. Since we only need the time to hit the ground (not the horizontal range), only the vertical component of the initial velocity and the height of the tower matter — the $30°$ launch angle only tells us how to split the speed into components.

   Step-by-Step Solution

   1. Vertical component of initial velocity: $u_y=50\sin30°=50\times0.5=25\ \mathrm{ms^{-1}}$ (upward).
   2. Take the top of the tower as origin, upward positive; ground is at displacement $y=-70\ \mathrm m$.
   3. Kinematics: $y=u_yt-\tfrac12gt^2 \Rightarrow -70=25t-5t^2$.
   4. Rearrange: $5t^2-25t-70=0 \Rightarrow t^2-5t-14=0$.
   5. Solve the quadratic: $t=\dfrac{5\pm\sqrt{25+56}}{2}=\dfrac{5\pm\sqrt{81}}{2}=\dfrac{5\pm9}{2}$. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.An object is projected from the top of a tower of height 'H' at angle $\theta$ with horizontal. It strikes the ground at P lying at a distance D from the foot of the tower. Calculate the maximum height attained by the object (A) $\dfrac{v^2 \sin^2\theta}{2g}$ (B) $H + \dfrac{D^2\tan^2\theta}{4(H+D\tan\theta)}$ (C) $H + D\tan\theta$ (D) $H + \dfrac{D^2}{v\cos\theta}$

   ›Reveal solutionSolution

   This tests eliminating the unknown launch speed $v$ from projectile kinematics using the known landing distance $D$, to express the maximum height purely in terms of $H$, $D$, and $\theta$. The answer is (B).

   Concept and Intuition

   The maximum height reached above the launch point in projectile motion is $\dfrac{(v\sin\theta)^2}{2g}$ — but here $v$ is not given directly. Instead we're given where the projectile lands (distance $D$ from the tower's foot), so we must first use the full trajectory equation to solve for $v^2\cos^2\theta$ in terms of known quantities, then substitute back.

   Step-by-Step Solution

   1. Place the origin at the foot of the tower. The projectile starts at height $H$ with velocity $v$ at angle $\theta$ above horizontal.
   2. Horizontal: $x=v\cos\theta\,t$. Vertical: $y=H+v\sin\theta\,t-\tfrac12 g t^2$.
   3. At landing, $x=D$ and $y=0$: from the horizontal equation, $t=\dfrac{D}{v\cos\theta}$.
   4. Substitute into the vertical equation: $0=H+v\sin\theta\cdot\dfrac{D}{v\cos\theta}-\tfrac12 g\left(\dfrac{D}{v\cos\theta}\right)^2=H+D\tan\theta-\dfrac{gD^2}{2v^2\cos^2\theta}$.
   5. Solve for $v^2\cos^2\theta$: $\dfrac{gD^2}{2v^2\cos^2\theta}=H+D\tan\theta \Rightarrow v^2\cos^2\theta=\dfrac{gD^2}{2(H+D\tan\theta)}$.
   6. Maximum height above the launch point is $\dfrac{(v\sin\theta)^2}{2g}=\dfrac{v^2\cos^2\theta\cdot\tan^2\theta}{2g}$ (using $v\sin\theta = v\cos\theta\tan\theta$). …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A cricketer can throw a ball with a speed of $V_0$. If he throws the ball, while running with a speed $V_0$, at an angle '$\alpha$' with the horizontal, then the range of the ball will be maximum when '$\alpha$' is (A) $45°$ (B) $30°$ (C) $60°$ (D) $53°$

   ›Reveal solutionSolution

   Because the cricketer's own running speed adds to the ball's horizontal component but not its vertical component, the range-maximizing angle shifts from the usual $45°$ down to $60°$... actually up — let's see: maximizing $\sin\alpha(1+\cos\alpha)$ gives $\alpha=60°$.

   Concept and Intuition

   When a projectile is launched from a moving platform, its ground-frame velocity is the vector sum of the platform's velocity and the launch velocity relative to the platform. Here the horizontal component gets boosted by the runner's speed while the vertical component is unaffected, so the range formula is no longer the simple $\frac{V_0^2\sin2\alpha}{g}$ and the optimal angle is not $45°$.

   Step-by-Step Solution

   1. Cricketer runs at speed $V_0$ (horizontal) and throws the ball at speed $V_0$ relative to himself, at angle $\alpha$ to the horizontal, in the plane of his run.
   2. Ground-frame velocity components of the ball: $v_x=V_0+V_0\cos\alpha=V_0(1+\cos\alpha)$ $v_y=V_0\sin\alpha$
   3. Time of flight (ball returns to same height): $T=\dfrac{2v_y}{g}=\dfrac{2V_0\sin\alpha}{g}$.
   4. Range: $R=v_x\cdot T=V_0(1+\cos\alpha)\cdot\dfrac{2V_0\sin\alpha}{g}=\dfrac{2V_0^2}{g}\sin\alpha(1+\cos\alpha)$.
   5. To maximize, differentiate $f(\alpha)=\sin\alpha+\sin\alpha\cos\alpha=\sin\alpha+\tfrac12\sin2\alpha$ with respect to $\alpha$: $f'(\alpha)=\cos\alpha+\cos2\alpha=0$. …

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6. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A ball of mass 100 g is projected with velocity $20\ \text{ms}^{-1}$ at $60^{\circ}$ with horizontal. The decrease in kinetic energy of the ball during its entire upward journey is (A) 15 J (B) 20 J (C) Zero (D) 5 J

   ›Reveal solutionSolution

   KE drops from 20 J (launch) to 5 J (top, only horizontal speed remains), a

   decrease of 15 J. Answer: (A).

   Concept and Intuition

   In projectile motion, the vertical velocity component continuously decreases

   under gravity and becomes zero exactly at the peak of the trajectory, while

   the horizontal component stays constant throughout (no horizontal force in

   ideal projectile motion). So the kinetic energy "lost" during the upward

   journey is entirely due to the loss of the vertical velocity component — it

   gets converted into gravitational potential energy.

   Step-by-Step Solution

   1. Initial speed $u = 20\ \text{ms}^{-1}$ at $60^\circ$ to horizontal; mass $m = 0.1$ kg $= 100$ g.
   2. Initial KE $= \tfrac12 m u^2 = \tfrac12(0.1)(400) = 20$ J.
   3. At the highest point, vertical velocity $=0$; only $u_x = u\cos60^\circ = 10\ \text{ms}^{-1}$ remains. …

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7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.An object is projected with an angle of $60°$ with horizontal with a velocity V. During the path, when it makes $30°$ with horizontal, its velocity becomes $10\ \mathrm{ms^{-1}}$, then V is (A) $10\sqrt{3}\ \mathrm{ms^{-1}}$ (B) $\sqrt{30}\ \mathrm{ms^{-1}}$ (C) $3\sqrt{10}\ \mathrm{ms^{-1}}$ (D) $\sqrt{3}\ \mathrm{ms^{-1}}$

   ›Reveal solutionSolution

   The horizontal velocity component is conserved in projectile motion; equating it at launch and at the $30°$ point gives $V=10\sqrt3\ \mathrm{ms^{-1}}$.

   Concept and Intuition

   In projectile motion (no air resistance), gravity acts only vertically, so the horizontal component of velocity, $v_x=v\cos\theta$, never changes during the flight — it's the same at launch as at any other point on the trajectory.

   Step-by-Step Solution

   1. At launch, the velocity is $V$ at $60°$ to horizontal, so the horizontal component is

   $v_x = V\cos60° = \frac{V}{2}$

   2. Later, when the velocity vector makes $30°$ with horizontal and has magnitude $10\ \mathrm{ms^{-1}}$, its horizontal component is

   $v_x = 10\cos30° = 10\cdot\frac{\sqrt3}{2}=5\sqrt3$

   3. Since $v_x$ is conserved throughout the flight: …

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8. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.As shown in figure, a particle slides on a frictionless track which terminates in a straight line horizontal section B. If the particle starts slipping from A, then the horizontal distance to be covered by the particle before it hits the ground after crossing B. [FIGURE] (a curved frictionless track: it starts at a raised point A at height $h_1 = 1\ m$ above the ground, dips down to ground level, then rises again to a raised horizontal section at point B at height $h_2 = \frac{1}{2}\ m$ above the ground; the track ends at B, after which the particle becomes a projectile) (A) $0.5\ m$ (B) $1\ m$ (C) $1.5\ m$ (D) $2\ m$

   ›Reveal solutionSolution

   This combines energy conservation on a frictionless track with horizontal projectile motion after the particle leaves the track at B — the answer is a clean $1\ m$.

   Concept and Intuition

   On the frictionless track, only gravity does work, so mechanical energy is conserved from A to B. This gives the speed at B purely from the height drop $h_1 - h_2$ (the shape of the track in between doesn't matter). Once the particle leaves the horizontal section at B, it becomes a projectile: it moves horizontally with the constant speed $v_B$ while falling freely under gravity from height $h_2$. The horizontal range is simply (horizontal speed) $\times$ (time to fall).

   Step-by-Step Solution

   1. Energy conservation, A to B (starting from rest at A): $\tfrac12 v_B^2 = g(h_1-h_2)$, so $v_B^2 = 2g(h_1-h_2) = 2(10)(1-0.5)=10\ \mathrm{m^2/s^2}$, giving $v_B=\sqrt{10}\ \mathrm{ms^{-1}}$.
   2. At B, the track is horizontal, so the particle leaves with velocity $v_B$ purely horizontal, at height $h_2 = 0.5\ m$ above the ground. …

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9. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.The percentage decrease in range of a projectile projected at $30°$ when compared to maximum range is (A) 13.4 % (B) 25 % (C) 50 % (D) 75 %

   ›Reveal solutionSolution

   Compare the range at $30^\circ$ to the maximum range (at $45^\circ$); the percentage decrease works out to $13.4\%$.

   Concept and Intuition

   For a projectile launched with speed $u$ at angle $\theta$, the horizontal range is $R=\dfrac{u^2\sin2\theta}{g}$. This is maximum when $\sin2\theta=1$, i.e. $\theta=45^\circ$, giving $R_{max}=u^2/g$.

   Step-by-Step Solution

   1. $R_{max}=\dfrac{u^2}{g}$ (at $\theta=45^\circ$).
   2. $R_{30}=\dfrac{u^2\sin60^\circ}{g}=\dfrac{u^2(\sqrt3/2)}{g}=\dfrac{\sqrt3}{2}R_{max}$.
   3. $\dfrac{\sqrt3}{2}\approx0.8660$. …

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10. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
    Q.A bomber plane moves horizontally with a speed of 500 m s$^{-1}$ and a bomb released from it strikes the ground in 10 sec. Angle with which it strikes the ground will be (g = 10 m s$^{-2}$) (A) $\tan^{-1}\left(\dfrac{1}{5}\right)$ (B) $\tan^{-1}\left(\dfrac{1}{2}\right)$ (C) $\tan^{-1}2$ (D) $\tan^{-1}5$

    ›Reveal solutionSolution

    Classic horizontal-projectile problem: the impact angle is found from the ratio of vertical to horizontal velocity components at landing. Answer: (A).

    Concept and Intuition

    A bomb released from a horizontally-moving plane undergoes projectile motion: the horizontal component of velocity never changes (no horizontal force acts, air resistance ignored), while the vertical component grows linearly under gravity from zero. The angle the resultant velocity makes with the horizontal (i.e., with the ground) at any instant is simply $\tan^{-1}(v_y/v_x)$.

    Step-by-Step Solution

    1. Horizontal velocity at impact: $v_x = u = 500\ \text{ms}^{-1}$ (unchanged).
    2. Vertical velocity at impact: $v_y = g t = 10 \times 10 = 100\ \text{ms}^{-1}$.
    3. Angle with the ground: $\tan\theta = v_y/v_x = 100/500 = 1/5$. …

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11. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If bullets are fired in all possible directions from same point with equal velocity of $10\ \mathrm{m\,s^{-1}}$ and with an angle of projection $45^\circ$, then the area covered by the bullets on the ground is nearly (Acceleration due to gravity $10\ \mathrm{m\,s^{-2}}$) (A) $628\ \mathrm{m^2}$ (B) $314\ \mathrm{m^2}$ (C) $157\ \mathrm{m^2}$ (D) $79\ \mathrm{m^2}$

    ›Reveal solutionSolution

    This tests recognizing that bullets fired at a fixed 45° elevation in all horizontal directions all land at the maximum range, tracing out a circle; the answer is (B).

    Concept and Intuition

    For projectile motion, range $R=\dfrac{u^2\sin2\theta}{g}$ depends only on the elevation angle $\theta$, not the azimuthal (compass) direction. If bullets are fired at $\theta=45^\circ$ in every horizontal direction from the same point, each one travels the same range $R$ (the maximum possible range, since $\sin90^\circ=1$), so all the landing points trace out a full circle of radius $R$. The interior of this circle (all points reachable) is the region "covered," and geometrically its boundary here is exactly the locus of landing points at this fixed angle.

    Step-by-Step Solution

    1. Range formula: $R=\dfrac{u^2\sin2\theta}{g}$.
    2. With $\theta=45^\circ$: $\sin90^\circ=1$, so $R=\dfrac{u^2}{g}$.
    3. Substitute $u=10\ \mathrm{m/s}$, $g=10\ \mathrm{m/s^2}$: $R=\dfrac{100}{10}=10$ m. …

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12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.A ball is projected from a point with a speed $V_0$ at certain angle with the horizontal. From the same point and at the same instant, a person starts running with a constant speed $0.5V_0$ to catch the ball. If the person catches the ball after some time, then the angle of projection of the ball is (A) $60^\circ$ (B) $30^\circ$ (C) $45^\circ$ (D) $53^\circ$

    ›Reveal solutionSolution

    This tests matching horizontal velocity components so a constant-speed runner can catch a projectile; the answer is (A).

    Concept and Intuition

    A person running at constant speed can only catch the ball if the ball's horizontal velocity component (which is constant throughout projectile motion, since there's no horizontal acceleration) exactly equals the runner's speed. If it didn't match, the horizontal gap between them would keep changing and could never close to zero except momentarily at the start.

    Step-by-Step Solution

    1. The ball is projected with speed $V_0$ at angle $\theta$. Its horizontal velocity component is $V_0\cos\theta$, and (crucially) this stays constant throughout the flight since there is no horizontal force.
    2. The person runs with constant speed $0.5V_0$ starting at the same instant and same point. …

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