📘 Motion In Two Dimensions — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Motion In Two Dimensions for JEE Main with previous year questions on projectile motion, circular motion, and relative motion. Each solution is explained in a simple, step-by-step manner to boost your score.
Motion In Two Dimensions — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (7 April Shift 1)
Two projectiles are fired from ground with same initial speeds from same point at angles \((45^\circ + \alpha)\) and \((45^\circ - \alpha)\) with horizontal direction. The ratio of their times of flights is :
(1) 1 (2) \(\frac{1-\tan\alpha}{1+\tan\alpha}\)
(3) \(\frac{1+\sin 2\alpha}{1-\sin 2\alpha}\) (4) \(\frac{1+\tan\alpha}{1-\tan\alpha}\)
(1) 1 (2) \(\frac{1-\tan\alpha}{1+\tan\alpha}\)
(3) \(\frac{1+\sin 2\alpha}{1-\sin 2\alpha}\) (4) \(\frac{1+\tan\alpha}{1-\tan\alpha}\)
✅ Answer & Explanation:
Correct Option: (4)
Time of flight \(T = \frac{2u \sin\theta}{g}\).
Ratio \(\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)} = \frac{\sin 45^\circ \cos\alpha + \cos 45^\circ \sin\alpha}{\sin 45^\circ \cos\alpha - \cos 45^\circ \sin\alpha}\)
Dividing by \(\cos 45^\circ \cos\alpha\):
Ratio \(= \frac{1 + \tan\alpha}{1 - \tan\alpha}\).
Time of flight \(T = \frac{2u \sin\theta}{g}\).
Ratio \(\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)} = \frac{\sin 45^\circ \cos\alpha + \cos 45^\circ \sin\alpha}{\sin 45^\circ \cos\alpha - \cos 45^\circ \sin\alpha}\)
Dividing by \(\cos 45^\circ \cos\alpha\):
Ratio \(= \frac{1 + \tan\alpha}{1 - \tan\alpha}\).
Q2. JEE Main 2025 (3 April Shift 2)
A particle is projected with velocity \(u\) so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as \(\frac{nu^2}{25g}\), where value of \(n\) is : (Given '\(g\)' is the acceleration due to gravity).
(1) 6 (2) 18
(3) 12 (4) 24
(1) 6 (2) 18
(3) 12 (4) 24
✅ Answer & Explanation:
Correct Option: (4)
Given \(R = 3H\).
\(\frac{u^2 \sin 2\theta}{g} = 3 \frac{u^2 \sin^2\theta}{2g} \implies 2 \sin\theta \cos\theta = \frac{3}{2} \sin^2\theta \implies \tan\theta = \frac{4}{3}\).
From \(\tan\theta = 4/3\), we find \(\sin\theta = 4/5\) and \(\cos\theta = 3/5\).
\(R = \frac{u^2 (2 \cdot \frac{4}{5} \cdot \frac{3}{5})}{g} = \frac{24u^2}{25g}\).
Thus, \(n = 24\).
Given \(R = 3H\).
\(\frac{u^2 \sin 2\theta}{g} = 3 \frac{u^2 \sin^2\theta}{2g} \implies 2 \sin\theta \cos\theta = \frac{3}{2} \sin^2\theta \implies \tan\theta = \frac{4}{3}\).
From \(\tan\theta = 4/3\), we find \(\sin\theta = 4/5\) and \(\cos\theta = 3/5\).
\(R = \frac{u^2 (2 \cdot \frac{4}{5} \cdot \frac{3}{5})}{g} = \frac{24u^2}{25g}\).
Thus, \(n = 24\).
Q3. JEE Main 2025 (3 April Shift 1)
The angle of projection of a particle is measured from the vertical axis as \(\phi\) and the maximum height reached by the particle is \(h_m\). Here \(h_m\) as function of \(\phi\) can be presented as :
Choose the correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (3)
If \(\phi\) is the angle from the vertical, the angle from the horizontal is \(\theta = 90^\circ - \phi\).
\(h_m = \frac{u^2 \sin^2\theta}{2g} = \frac{u^2 \sin^2(90^\circ - \phi)}{2g} = \frac{u^2 \cos^2\phi}{2g}\).
At \(\phi = 0^\circ\), \(h_m\) is maximum. At \(\phi = 90^\circ\), \(h_m = 0\). Graph (3) correctly represents this \(\cos^2\phi\) relationship.
If \(\phi\) is the angle from the vertical, the angle from the horizontal is \(\theta = 90^\circ - \phi\).
\(h_m = \frac{u^2 \sin^2\theta}{2g} = \frac{u^2 \sin^2(90^\circ - \phi)}{2g} = \frac{u^2 \cos^2\phi}{2g}\).
At \(\phi = 0^\circ\), \(h_m\) is maximum. At \(\phi = 90^\circ\), \(h_m = 0\). Graph (3) correctly represents this \(\cos^2\phi\) relationship.
Q4. JEE Main 2025 (29 Jan Shift 1)
Two projectiles are fired with same initial speed from same point on ground at angles of \((45^\circ - \alpha)\) and \((45^\circ + \alpha)\), respectively, with the horizontal direction. The ratio of their maximum heights attained is :
(1) \(\frac{1-\tan\alpha}{1+\tan\alpha}\) (2) \(\frac{1-\sin 2\alpha}{1+\sin 2\alpha}\)
(3) \(\frac{1+\sin 2\alpha}{1-\sin 2\alpha}\) (4) \(\frac{1+\sin\alpha}{1-\sin\alpha}\)
(1) \(\frac{1-\tan\alpha}{1+\tan\alpha}\) (2) \(\frac{1-\sin 2\alpha}{1+\sin 2\alpha}\)
(3) \(\frac{1+\sin 2\alpha}{1-\sin 2\alpha}\) (4) \(\frac{1+\sin\alpha}{1-\sin\alpha}\)
✅ Answer & Explanation:
Correct Option: (2)
\(H \propto \sin^2\theta\).
Ratio \(= \frac{\sin^2(45^\circ - \alpha)}{\sin^2(45^\circ + \alpha)} = \left[ \frac{\sin 45^\circ \cos\alpha - \cos 45^\circ \sin\alpha}{\sin 45^\circ \cos\alpha + \cos 45^\circ \sin\alpha} \right]^2 = \left[ \frac{1 - \tan\alpha}{1 + \tan\alpha} \right]^2\)
Using the identity \(\left(\frac{1-\tan\alpha}{1+\tan\alpha}\right)^2 = \frac{1-\sin 2\alpha}{1+\sin 2\alpha}\).
\(H \propto \sin^2\theta\).
Ratio \(= \frac{\sin^2(45^\circ - \alpha)}{\sin^2(45^\circ + \alpha)} = \left[ \frac{\sin 45^\circ \cos\alpha - \cos 45^\circ \sin\alpha}{\sin 45^\circ \cos\alpha + \cos 45^\circ \sin\alpha} \right]^2 = \left[ \frac{1 - \tan\alpha}{1 + \tan\alpha} \right]^2\)
Using the identity \(\left(\frac{1-\tan\alpha}{1+\tan\alpha}\right)^2 = \frac{1-\sin 2\alpha}{1+\sin 2\alpha}\).
Q5. JEE Main 2025 (22 Jan Shift 1)
A particle is projected at an angle of \(30^\circ\) from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is \(h_0\) and height traversed in the last second, before it reaches the maximum height, is \(h_1\). The ratio \(h_0 : h_1\) is _________ [Take, \(g = 10 \, \text{m/s}^2\)]
✅ Answer & Explanation:
Correct Answer: 5
Vertical velocity \(u_y = 60 \sin 30^\circ = 30 \, \text{m/s}\).
Time to reach max height \(t_{max} = u_y / g = 30 / 10 = 3 \, \text{s}\).
Height in 1st second: \(h_0 = 30(1) - \frac{1}{2}(10)(1)^2 = 25 \, \text{m}\).
Height in last second (from \(t=2\) to \(t=3\)):
\(h(3) = 30(3) - 5(9) = 45 \, \text{m}\).
\(h(2) = 30(2) - 5(4) = 40 \, \text{m}\).
\(h_1 = 45 - 40 = 5 \, \text{m}\).
Ratio \(h_0 / h_1 = 25 / 5 = 5\).
Vertical velocity \(u_y = 60 \sin 30^\circ = 30 \, \text{m/s}\).
Time to reach max height \(t_{max} = u_y / g = 30 / 10 = 3 \, \text{s}\).
Height in 1st second: \(h_0 = 30(1) - \frac{1}{2}(10)(1)^2 = 25 \, \text{m}\).
Height in last second (from \(t=2\) to \(t=3\)):
\(h(3) = 30(3) - 5(9) = 45 \, \text{m}\).
\(h(2) = 30(2) - 5(4) = 40 \, \text{m}\).
\(h_1 = 45 - 40 = 5 \, \text{m}\).
Ratio \(h_0 / h_1 = 25 / 5 = 5\).
Q6. JEE Main 2023 (08 Apr Shift 2)
The trajectory of projectile, projected from the ground is given by \(y = x - \frac{x^2}{20}\). Where \(x\) and \(y\) are measured in meter. The maximum height attained by the projectile will be :
(1) 200 m (2) 10 m
(3) 5 m (4) \(10\sqrt{2} \, \text{m}\)
(1) 200 m (2) 10 m
(3) 5 m (4) \(10\sqrt{2} \, \text{m}\)
✅ Answer & Explanation:
Correct Option: (3)
Comparing with \(y = x \tan\theta - \frac{gx^2}{2u^2 \cos^2\theta}\):
\(\tan\theta = 1 \implies \theta = 45^\circ\).
Range \(R\) is where \(y=0\): \(x(1 - x/20) = 0 \implies R = 20 \, \text{m}\).
For a projectile, \(H = \frac{R \tan\theta}{4} = \frac{20 \times 1}{4} = 5 \, \text{m}\).
Comparing with \(y = x \tan\theta - \frac{gx^2}{2u^2 \cos^2\theta}\):
\(\tan\theta = 1 \implies \theta = 45^\circ\).
Range \(R\) is where \(y=0\): \(x(1 - x/20) = 0 \implies R = 20 \, \text{m}\).
For a projectile, \(H = \frac{R \tan\theta}{4} = \frac{20 \times 1}{4} = 5 \, \text{m}\).
Q7. JEE Main 2022 (29 Jul Shift 1)
An object is projected in the air with initial velocity \(u\) at an angle \(\theta\). The projectile motion is such that the horizontal range \(R\), is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be _____ degree.
✅ Answer & Explanation:
Correct Answer: 15
Max range \(R = u^2/g\) occurs at \(\theta = 45^\circ\).
New range \(R' = R/2 = \frac{u^2 \sin 2\theta'}{g}\).
\(\frac{1}{2} \left(\frac{u^2}{g}\right) = \frac{u^2 \sin 2\theta'}{g} \implies \sin 2\theta' = \frac{1}{2} \implies 2\theta' = 30^\circ \implies \theta' = 15^\circ\).
Max range \(R = u^2/g\) occurs at \(\theta = 45^\circ\).
New range \(R' = R/2 = \frac{u^2 \sin 2\theta'}{g}\).
\(\frac{1}{2} \left(\frac{u^2}{g}\right) = \frac{u^2 \sin 2\theta'}{g} \implies \sin 2\theta' = \frac{1}{2} \implies 2\theta' = 30^\circ \implies \theta' = 15^\circ\).
Q8. JEE Main 2022 (26 Jul Shift 1)
If the initial velocity in horizontal direction of a projectile is unit vector \(\hat{i}\) and the equation of trajectory is \(y = 5x(1 - x)\). The \(y\) component vector of the initial velocity is _____ \(\hat{j}\). (Take \(g = 10 \, \text{m/s}^2\))
✅ Answer & Explanation:
Correct Answer: 5
Equation of trajectory: \(y = 5x - 5x^2\).
The slope at projection (\(x=0\)) is \(\frac{dy}{dx} = 5 - 10x\). At \(x=0\), slope \(= 5\).
Also, slope \(= \frac{u_y}{u_x}\). Since \(u_x = 1\), then \(u_y = 5\).
Equation of trajectory: \(y = 5x - 5x^2\).
The slope at projection (\(x=0\)) is \(\frac{dy}{dx} = 5 - 10x\). At \(x=0\), slope \(= 5\).
Also, slope \(= \frac{u_y}{u_x}\). Since \(u_x = 1\), then \(u_y = 5\).
Q9. JEE Main 2025 (24 Jan Shift 2)
The position vector of a moving body at any instant of time is given as \(\vec{r} = (5t^2 \hat{i} - 5t \hat{j}) \, \text{m}\). The magnitude and direction of velocity at \(t = 2 \, \text{s}\) is,
(1) \(5\sqrt{15} \, \text{m/s}\), making an angle of \(\tan^{-1} 4\) with -ve Y axis
(2) \(5\sqrt{15} \, \text{m/s}\), making an angle of \(\tan^{-1} 4\) with +ve X axis
(3) \(5\sqrt{17} \, \text{m/s}\), making an angle of \(\tan^{-1} 4\) with +ve X axis
(4) \(5\sqrt{17} \, \text{m/s}\), making an angle of \(\tan^{-1} 4\) with -ve Y axis
(1) \(5\sqrt{15} \, \text{m/s}\), making an angle of \(\tan^{-1} 4\) with -ve Y axis
(2) \(5\sqrt{15} \, \text{m/s}\), making an angle of \(\tan^{-1} 4\) with +ve X axis
(3) \(5\sqrt{17} \, \text{m/s}\), making an angle of \(\tan^{-1} 4\) with +ve X axis
(4) \(5\sqrt{17} \, \text{m/s}\), making an angle of \(\tan^{-1} 4\) with -ve Y axis
✅ Answer & Explanation:
Correct Option: (4)
\(\vec{v} = \frac{d\vec{r}}{dt} = 10t \hat{i} - 5 \hat{j}\).
At \(t = 2 \, \text{s}\), \(\vec{v} = 20 \hat{i} - 5 \hat{j} \, \text{m/s}\).
Magnitude \(= \sqrt{20^2 + (-5)^2} = \sqrt{425} = 5\sqrt{17} \, \text{m/s}\).
Direction: Let \(\alpha\) be the angle with the -ve Y-axis.
\(\tan\alpha = \frac{v_x}{|v_y|} = \frac{20}{5} = 4 \implies \alpha = \tan^{-1} 4\).
\(\vec{v} = \frac{d\vec{r}}{dt} = 10t \hat{i} - 5 \hat{j}\).
At \(t = 2 \, \text{s}\), \(\vec{v} = 20 \hat{i} - 5 \hat{j} \, \text{m/s}\).
Magnitude \(= \sqrt{20^2 + (-5)^2} = \sqrt{425} = 5\sqrt{17} \, \text{m/s}\).
Direction: Let \(\alpha\) be the angle with the -ve Y-axis.
\(\tan\alpha = \frac{v_x}{|v_y|} = \frac{20}{5} = 4 \implies \alpha = \tan^{-1} 4\).
Q10. JEE Main 2025 (2 April Shift 1)
A river is flowing from west to east direction with speed of \(9 \, \text{km/h}\). If a boat capable of moving at a maximum speed of \(27 \, \text{km/h}\) in still water, crosses the river in half a minute, while moving with maximum speed at an angle of \(150^\circ\) to direction of river flow, then the width of the river is :
(1) 300 m (2) 112.5 m
(3) 75 m (4) \(112.5 \times \sqrt{3} \, \text{m}\)
(1) 300 m (2) 112.5 m
(3) 75 m (4) \(112.5 \times \sqrt{3} \, \text{m}\)
✅ Answer & Explanation:
Correct Option: (2)
Let river be along +X axis. Boat velocity relative to river makes \(150^\circ\) with river flow.
Vertical component (crossing velocity) \(v_y = 27 \sin 150^\circ = 27 \times \frac{1}{2} = 13.5 \, \text{km/h}\).
Time \(t = 0.5 \, \text{min} = \frac{1}{120} \, \text{hour}\).
Width \(w = v_y \times t = 13.5 \times \frac{1}{120} = 0.1125 \, \text{km} = 112.5 \, \text{m}\).
Let river be along +X axis. Boat velocity relative to river makes \(150^\circ\) with river flow.
Vertical component (crossing velocity) \(v_y = 27 \sin 150^\circ = 27 \times \frac{1}{2} = 13.5 \, \text{km/h}\).
Time \(t = 0.5 \, \text{min} = \frac{1}{120} \, \text{hour}\).
Width \(w = v_y \times t = 13.5 \times \frac{1}{120} = 0.1125 \, \text{km} = 112.5 \, \text{m}\).
Q11. JEE Main 2022 (27 Jun Shift 1)
A girl standing on road holds her umbrella at \(45^\circ\) with the vertical to keep the rain away. If she starts running without umbrella with a speed of \(15\sqrt{2} \, \text{km/h}\), the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is :
(1) \(30 \, \text{km/h}\) (2) \(\frac{25}{\sqrt{2}} \, \text{km/h}\)
(3) \(\frac{30}{\sqrt{2}} \, \text{km/h}\) (4) \(25 \, \text{km/h}\)
(3) \(\frac{30}{\sqrt{2}} \, \text{km/h}\) (4) \(25 \, \text{km/h}\)
✅ Answer & Explanation:
Correct Option: (3)
Let speed of rain wrt girl be \(v\). Then \(\vec{v}_{rg} = -v \hat{j}\).
Girl's speed \(\vec{v}_g = 15\sqrt{2} \hat{i}\).
Rain speed wrt ground \(\vec{v}_r = \vec{v}_{rg} + \vec{v}_g = 15\sqrt{2} \hat{i} - v \hat{j}\).
Given initial angle is \(45^\circ\): \(\tan 45^\circ = \frac{|v_{rx}|}{|v_{ry}|} = \frac{15\sqrt{2}}{v} = 1 \implies v = 15\sqrt{2} \, \text{km/h}\).
Note: \(15\sqrt{2} = \frac{30}{\sqrt{2}}\).
Let speed of rain wrt girl be \(v\). Then \(\vec{v}_{rg} = -v \hat{j}\).
Girl's speed \(\vec{v}_g = 15\sqrt{2} \hat{i}\).
Rain speed wrt ground \(\vec{v}_r = \vec{v}_{rg} + \vec{v}_g = 15\sqrt{2} \hat{i} - v \hat{j}\).
Given initial angle is \(45^\circ\): \(\tan 45^\circ = \frac{|v_{rx}|}{|v_{ry}|} = \frac{15\sqrt{2}}{v} = 1 \implies v = 15\sqrt{2} \, \text{km/h}\).
Note: \(15\sqrt{2} = \frac{30}{\sqrt{2}}\).
Q12. JEE Main 2021 (27 Jul Shift 2)
A swimmer wants to cross a river from point A to point B. Line AB makes an angle of \(30^\circ\) with the flow of the river. The magnitude of the velocity of the swimmer is the same as that of the river. The angle \(\theta\) with the line AB should be ______ \(^\circ\), so that the swimmer reaches point B.
✅ Answer & Explanation:
Correct Answer: 30
For the resultant velocity to be along AB, the component of river velocity and swimmer velocity perpendicular to AB must cancel.
Perpendicular component of river velocity \(= u \sin 30^\circ\).
Perpendicular component of swimmer velocity \(= v \sin \theta\).
Since \(v = u\), then \(\sin\theta = \sin 30^\circ \implies \theta = 30^\circ\).
For the resultant velocity to be along AB, the component of river velocity and swimmer velocity perpendicular to AB must cancel.
Perpendicular component of river velocity \(= u \sin 30^\circ\).
Perpendicular component of swimmer velocity \(= v \sin \theta\).
Since \(v = u\), then \(\sin\theta = \sin 30^\circ \implies \theta = 30^\circ\).
Q13. JEE Main 2021 (24 Feb Shift 2)
A particle is projected with velocity \(v_0\) along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. \(ma = -\alpha x^2\). The distance at which the particle stops :
(1) \(\left(\frac{2v_0}{3\alpha}\right)^{1/3}\) (2) \(\left(\frac{3mv_0^2}{2\alpha}\right)^{1/3}\)
(3) \(\left(\frac{3v_0^2}{2\alpha}\right)^{1/2}\) (4) \(\left(\frac{2v_0^2}{3\alpha}\right)^{1/2}\)
(1) \(\left(\frac{2v_0}{3\alpha}\right)^{1/3}\) (2) \(\left(\frac{3mv_0^2}{2\alpha}\right)^{1/3}\)
(3) \(\left(\frac{3v_0^2}{2\alpha}\right)^{1/2}\) (4) \(\left(\frac{2v_0^2}{3\alpha}\right)^{1/2}\)
✅ Answer & Explanation:
Correct Option: (2)
\(m v \frac{dv}{dx} = -\alpha x^2 \implies \int_{v_0}^{0} v dv = -\frac{\alpha}{m} \int_{0}^{s} x^2 dx\)
\(- \frac{v_0^2}{2} = -\frac{\alpha}{m} \left( \frac{s^3}{3} \right) \implies s^3 = \frac{3mv_0^2}{2\alpha}\)
\(s = \left(\frac{3mv_0^2}{2\alpha}\right)^{1/3}\).
\(m v \frac{dv}{dx} = -\alpha x^2 \implies \int_{v_0}^{0} v dv = -\frac{\alpha}{m} \int_{0}^{s} x^2 dx\)
\(- \frac{v_0^2}{2} = -\frac{\alpha}{m} \left( \frac{s^3}{3} \right) \implies s^3 = \frac{3mv_0^2}{2\alpha}\)
\(s = \left(\frac{3mv_0^2}{2\alpha}\right)^{1/3}\).
📘 Exam Preparation Tip for Motion in Two Dimensions
- Master Projectile Motion: Memorize the key formulas: Time of flight \(T = \frac{2u\sin\theta}{g}\), Range \(R = \frac{u^2\sin 2\theta}{g}\), and Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\). Understand how these are derived.
- Trajectory Equation: The equation \(y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}\) is powerful. Use it to find range, maximum height, and other parameters without memorizing separate formulas.
- Relative Motion in 2D: For river-boat and rain-man problems, break velocities into components. The key is that the velocity of the boat relative to the river or rain relative to the girl is the vector difference of their velocities.
- Graph Interpretation: Practice interpreting \(h_m\) vs \(\phi\) and other graphs. Understand how the variation of quantities with the angle of projection follows trigonometric functions.
- Use Calculus for Variable Acceleration: For problems like Q54, where acceleration is a function of displacement, use \(a = v\frac{dv}{dx}\) and integrate to find the stopping distance.