Motion In Two Dimensions JEE Main previous year questions with solutions

5 solved JEE Main questions on Motion In Two Dimensions, free to read — no sign-in needed. The full chapter has 127 questions; sign in to attempt the remaining 122 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Projectile motion
   Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5$ s and $10$ s, respectively, then the value of $R$ is ________ m. (Take $g=10$ m/s$^2$)
   1. A.$250$
   2. B.$25$
   3. C.$500$
   4. D.$125$
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   Answer: (A)

   For two projectiles to have the same horizontal range $R$ with the same initial speed $u$, their angles of projection must be complementary, i.e., $\theta$ and $90^\circ - \theta$. The times of flight for these two angles are given by: $t_1 = \dfrac{2u \sin \theta}{g}$ $t_2 = \dfrac{2u \cos \theta}{g}$ Multiplying $t_1$ and $t_2$, we get: $t_1 t_2 = \left(\dfrac{2u \sin \theta}{g}\right) \left(\dfrac{2u \cos \theta}{g}\right) = \dfrac{2}{g} \left(\dfrac{u^2 (2 \sin \theta \cos \theta)}{g}\right)$ Since the horizontal range is $R = \dfrac{u^2 \sin 2\theta}{g}$, we can write: $t_1 t_2 = \dfrac{2R}{g}$ Given $t_1 = 5$ s, $t_2 = 10$ s, and $g = 10$ m/s$^2$, substituting these values: $5 \times 10 = \dfrac{2R}{10}$ $50 = \dfrac{2R}{10}$ $2R = 500 \Rightarrow R = 250$ m Answer: $250$
2. Q2JEE Main 2024 (05 Apr, Shift 2)Uniform Circular Motion
   A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 \mathrm{~m}$ and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is $\left(\right.$ in $\mathrm{m} / \mathrm{s}^2$ ) :
   1. A.$57600 \pi^2 \mathrm{~ms}^{-2}$
   2. B.Zero
   3. C.$4 \pi^2 \mathrm{~ms}^{-2}$
   4. D.$16 \pi^2 \mathrm{~ms}^{-2}$
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   Answer: (D)

   $\begin{aligned} & \text { Given : } \mathrm{R}=9 \mathrm{~m}, \\ & 120 \text { revolution in } 3 \mathrm{~min} \\ & \omega=\frac{120 \mathrm{Rev} .}{3 \mathrm{~min} .}=\frac{120 \times 2 \pi \mathrm{rad}}{3 \times 60 \mathrm{sec}}=\frac{4 \pi}{3} \mathrm{rad} / \mathrm{s} \\ & \mathrm{a}_{\text {centripetal }}=\omega^2 \mathrm{R}=\left(\frac{4 \pi}{3}\right)^2 \times 9=16 \pi^2 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
3. Q3JEE Main 2022 (28 Jul, Shift 2)Relative motion
   At time $t=0$ a particle starts travelling from a height $7\hat{z}cm$ in a plane keeping $z$ coordinate constant. At any instant of time, it's position along the $x$ and $y$ directions are defined as $3t$ and $5{t}^{3}$ respectively. At $t=1s$ acceleration of the particle will be
   1. A.$-30y$
   2. B.$30y$
   3. C.$3x+15y$
   4. D.$3x+15y+7\hat{z}$
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   Answer: (B)

   The position vector of the particle can be written as, $\vec{r}=3t\hat{i}+5{t}^{3}\hat{j}+7\hat{k}$ The velocity of the particle will be, $\vec{v}=\dfrac{d\vec{r}}{dt}=3\hat{i}+15{t}^{2}\hat{j}$ Now the acceleration of the particle will be, $\vec{a}=\dfrac{d\vec{v}}{dt}=\dfrac{{d}^{2}\vec{r}}{d{t}^{2}}=30t\hat{j}$ At $t=1s\Rightarrow \dfrac{{d}^{2}r}{d{t}^{2}}=30\hat{j}$
4. Q4JEE Main 2026 (04 Apr, Shift 2)Projectile motion
   A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is $6.4$ m. The speed of the bullets from the gun is ______ m/s. (take $g=10$ m/s$^2$)
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   Answer: 8

   The maximum horizontal range of a projectile is given by $R_{\text{max}} = \dfrac{u^2}{g}$. Given $R_{\text{max}} = 6.4$ m and $g = 10$ m/s$^2$. Substituting the values: $6.4 = \dfrac{u^2}{10}$ $u^2 = 64$ $u = 8$ m/s. Answer: $8$
5. Q5JEE Main 2026 (04 Apr, Shift 2)Projectile motion
   If $x$ and $y$ coordinates of a projectile as a function of time $(t)$ are given as $24t$ and $43.6t-4.9t^2$, respectively, then the angle (in degrees) made by the projectile with horizontal when $t=2$ s is ______.
   1. A.$60$
   2. B.$45$
   3. C.$30$
   4. D.$75$
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   Answer: (B)

   The velocity components of the projectile are given by the derivatives of the position coordinates with respect to time. $v_x = \dfrac{dx}{dt} = \dfrac{d}{dt}(24t) = 24$ $v_y = \dfrac{dy}{dt} = \dfrac{d}{dt}(43.6t - 4.9t^2) = 43.6 - 9.8t$ At $t = 2$ s, the velocity components are: $v_x = 24$ $v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24$ The angle $\theta$ made by the velocity vector with the horizontal is given by: $\tan \theta = \dfrac{v_y}{v_x} = \dfrac{24}{24} = 1$ $\theta = 45^{\circ}$ Answer: $45$