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 104 questions; sign in to attempt the remaining 99 in the exam simulator.

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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 2025 (08 Apr, Shift 2)Non-uniform Motion
   A body of mass 2 kg moving with velocity of $\overrightarrow{\mathrm{v}}_{\mathrm{in}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \mathrm{ms}^{-1}$ enters into a constant force field of 6 N directed along positive z -axis. If the body remains in the field for a period of $\frac{5}{3}$ seconds, then velocity of the body when it emerges from force field is
   1. A.$4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
   2. B.$3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
   3. C.$3 \hat{i}+4 \hat{j}-5 \hat{k}$
   4. D.$3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\sqrt{5} \hat{\mathrm{k}}$
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   Answer: (B)

   $\begin{aligned} & \overrightarrow{\mathrm{a}}=\frac{\mathrm{B}}{2} \hat{\mathrm{k}}=3 \hat{\mathrm{k}}, \mathrm{t}=\frac{5}{3} \mathrm{~s} \\ & \overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{a}} \mathrm{t}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\end{aligned}$
3. Q3JEE Main 2023 (01 Feb, Shift 2)Gravitational potential and Potential Energy
   For a body projected at an angle with the horizontal from the ground, choose the correct statement
   1. A.Gravitational potential energy is maximum at the highest point.
   2. B.The horizontal component of velocity is zero at highest point.
   3. C.The vertical component of momentum is maximum at the highest point.
   4. D.The kinetic energy (K.E.) is zero at the highest point of projectile motion.
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   Answer: (A)

   In a projectile motion, a particle has two components of velocity. The horizontal component of velocity remains constant over time, while the vertical component of velocity changes. At highest point, the vertical component of velocity $\left({v}_{y}\right)$ becomes zero, but the horizontal component is non-zero. So the kinetic energy of the particle is not zero. The formula to calculate the potential energy $\left(U\right)$ of the particle is given by $U=mgh...................(1)$ where, $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is the vertical height attained by the object. Hence, the potential energy is maximum when the projectile attains the maximum height.
4. Q4JEE 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 $t=1s\Rightarrow \dfrac{{d}^{2}r}{d{t}^{2}}=30\hat{j}$
5. Q5JEE Main 2020 (06 Sep, Shift 2)Rest and Motion
   A particle moving in the $xy$-plane experiences a velocity dependent force $\vec{F}=k\left({υ}_{y}\hat{i}+{υ}_{x}\hat{j}\right)$, where ${υ}_{x}$ and ${υ}_{y}$ are the $x$ and $y$ components of its velocity $\vec{υ}$. If $\vec{a}$ is the acceleration of the particle, then which of the following statements is true for the particle ?
   1. A.quantity $\vec{υ}\times \vec{a}$ is constant in time
   2. B.$\vec{F}$ arises due to a magnetic field
   3. C.kinetic energy of particle is constant in time
   4. D.quantity $\vec{υ}.\vec{a}$ is constant in time
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   Answer: (A)

   $\vec{F}=k\vec{v}$ $\Rightarrow$ $\vec{F}‖\vec{v}$ $\Rightarrow$ $\vec{a}‖\vec{v}$ $\Rightarrow$ $\vec{v}\times \vec{a}=0$ always