Kinematics JEE Main Questions with Solutions, PYQs & Practice Problems

JEE Main and JEE Advanced Kinematics Weightage

Based on JEE Main previous year trends and analysis, the number of questions asked from Kinematics varies year to year. On the basis of JEE Main 2026 exam analysis, the weightage of kinematics is 4.42%.

Kinematics Last 10 Years Most Asked Topics

JEE Main kinematics questions asked in the last 10 years include different topics. The table below shows the most asked JEE Main and JEE Advanced kinematics question topics.

Also Check: JEE Main 2027 Study Guide Complete Notes, Important Concepts, Formulae and Practice Questions

JEE Main 2026 Kinematics Most Asked Concepts

The table below shows the Kinematics JEE Main questions that were asked in 2026. This table covers the number of questions asked in both the January and April sessions from different kinematics concepts.

Also check: How to Prepare for JEE Main 2027?

Important Kinematics Formulas

Before you solve Kinematics JEE Mains PYQs, it’s important learn the basic Kinematics formulas. These formulae are used quite often in numericals.

1. Average Speed $=\frac{\text { Total Distance }}{\text { Total Time }}$

2. Average Velocity $=\frac{\text { Displacement }}{\text { Total Time }}$

3. Equation of Motion:

• $v=u+a t$

• $s=u t+\frac{1}{2} a t^2$

• $v^2=u^2+2 a s$

4. Maximum Height of Projectile: $H=\frac{u^2 \sin ^2 \theta}{2 g}$

5. Time of Flight: $T=\frac{2 u \sin \theta}{g}$

6. Horizontal Range: $R=\frac{u^2 \sin 2 \theta}{q}$

Also Check: JEE Main 2027 Important Formulas

JEE Main and JEE Advanced Kinematics Previous Year Questions

Practising Kinematics JEE Main PYQ is one of the best ways to prepare for JEE Main 2027. The questions below will help understand the exam pattern, the key topics, and also the general level of difficulty.

Question 1: 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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

(1) 250

(2) 25

(3) 500

(4) 125

Correct Answer: (1)

Solution

For range to be same, angle of projection must be complementary angles.

$
\begin{aligned}
& T_1=\frac{2 u \sin \theta}{g}=10, T_2=\frac{2 u \cos \theta}{g}=5 \\
& \text { Range }=\frac{2(u \sin \theta)(u \cos \theta)}{g} \\
& =\frac{2(50)(25)}{10} m=250 m
\end{aligned}
$

Question 2: A new unit $(\alpha)$ of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of $\alpha$ units if light takes 6 min. 40 s to cover this distance?

(1) $200 \alpha$

(2) $400 \alpha$

(3) $300 \alpha$

(4) $500 \alpha$

Correct Answer: (2)

Solution

$\begin{aligned} & \alpha=3 \times 10^8 \mathrm{~m} \\ & \text { Distance }=(400) \times 3 \times 10^8 \mathrm{~m}=400 \alpha\end{aligned}$

Question 3: A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle $\theta_1$ and in the next 2 s it rotates through an angle $\theta_2$. The ratio $\frac{\theta_2}{\theta_1}$ is $\_\_\_\_$ .

(1) 6

(2) 3

(3) 4

(4) $\frac{1}{3}$

Correct Answer: (2)

Solution

Let $\alpha$ be angular acceleration
In first 2 sec

$\theta_1=\frac{1}{2} \cdot \alpha(2)^2=2 \alpha$
Angular covered in 4 sec

$\theta^{\prime}=\frac{1}{2}(\alpha)(4)^2=8 \alpha$

Angle in next 2 sec

$\begin{aligned}
& \theta_2=\theta^{\prime}-\theta_1=8 \alpha-2 \alpha \\
& \theta_2=6 \alpha \\
& \frac{\theta_2}{\theta_1}=\frac{6 \alpha}{2 \alpha}=3
\end{aligned}$

Question 4: From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is $\_\_\_\_$ m.

(Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ and neglect the air resistance)

(1) 13

(2) 11

(3) 10

(4) 12

Correct Answer: (1)

Solution:

At any time t after it is released velocity $\mathrm{V}=\mathrm{gt}$
Acceleration of the ball remains constant $=g$
∴ Both will be equal at $\mathrm{t}=1 \mathrm{sec}$
∴ Position of ball from ground will be $(18-5) \mathrm{m}$

= 13m

Question 5: 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 $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$. (take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

(1) 8

(2) 7

(3) 6

(4) 5

Correct Answer: (1)

Solution:

$\begin{aligned}
& R_{\max }=\frac{v^2}{g}=6.4 \\
& v=\sqrt{64}=8 \mathrm{~m} / \mathrm{s}
\end{aligned}$

Question 6: If $x$ and $y$ coordinates of a projectile as a function of time (t) are given as $24 t$ and $43.6 t-4.9 t^2$, respectively, then the angle (in degrees) made by the projectile with horizontal when $\mathrm{t}=2 \mathrm{~s}$ is $\_\_\_\_$ .

(1) 60

(2) 45

(3) 30

(4) 75

Correct Answer: (2)

Solution:

$\begin{aligned} & x=24 t \quad y=43.6 t-4.9 t^2 \\ & v_x=\frac{d x}{d t}=24 \\ & v_y=\frac{d y}{d t}=43.6-9.8 t \\ & \tan \theta=\frac{v_y}{v_x}=\frac{43.6-9.8 t}{24} \\ & \text { At } t=2 \quad \frac{43.6-19.6}{24} \\ & \tan \theta=1 \\ & \theta=45^{\circ}\end{aligned}$

Question 7: Two cars A and B are moving in the same direction along a straight line with speeds $100 \mathrm{~km} / \mathrm{h}$ and $80 \mathrm{~km} / \mathrm{h}$, respectively such that car A is moving ahead of car B. A person in car B throws a stone with a speed v so that it hits the car A with a speed of $5 \mathrm{~m} / \mathrm{s}$. The value of v is

$\_\_\_\_$ km/h.

(1) 18

(2) 28

(3) 38

(4) 48

Correct Answer: (3)

Solution:

$\begin{aligned} & \mathrm{V}=\overrightarrow{\mathrm{V}}_{\mathrm{SB}} \\ & \mathrm{V}=\overrightarrow{\mathrm{V}}_{\mathrm{S}}-\overrightarrow{\mathrm{V}}_{\mathrm{B}} \\ & \mathrm{V}_{\mathrm{B}}+\mathrm{V}=\overrightarrow{\mathrm{V}}_{\mathrm{S}} \\ & \overrightarrow{\mathrm{V}}_{\mathrm{S}}=\mathrm{V}+80 \\ & \overrightarrow{\mathrm{~V}}_{\mathrm{SA}}=5 \mathrm{~m} / \mathrm{s}=18 \mathrm{~km} / \mathrm{hr} \\ & \overrightarrow{\mathrm{V}}_{\mathrm{S}}-\overrightarrow{\mathrm{V}}_{\mathrm{A}}=\overrightarrow{\mathrm{V}}_{\mathrm{SA}} \\ & \mathrm{V}+80-100=18 \mathrm{~km} / \mathrm{hr} \\ & \mathrm{V}-20=18 \\ & \mathrm{~V}=38 \mathrm{~km} / \mathrm{hr}\end{aligned}$

Question 8: The two projectiles are projected with the same initial velocities at the $15^{\circ}$ and $30^{\circ}$ with respect to the horizontal. The ratio of their ranges is $1: x$. The value of $x$ is

(1) $\sqrt{2}$

(2) $\sqrt{3}$

(3) $2 \sqrt{3}$

(4) $\frac{1}{\sqrt{2}}$

Correct Answer: (2)

Solution:

$\begin{aligned} & R_1=\frac{u^2 \sin 2 \times 15^{\circ}}{g} \\ & R_2=\frac{u^2 \sin 2 \times 30^{\circ}}{g} \\ & \frac{R_1}{R_2}=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1}{\sqrt{3}}=\frac{1}{x} \\ & x=\sqrt{3}\end{aligned}$

Question 9: The velocity of a particle is given as $\vec{v}=-x \hat{i}+2 y \hat{j}-z \hat{k} \mathrm{~m} / \mathrm{s}$. The magnitude of acceleration at point ( $1,2,4$ ) is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}^2$.

(1) $\sqrt{6}$

(2) 9

(3) $\sqrt{33}$

(4) 0

Correct Answer: (2)

Solution:

$\begin{aligned} & \vec{v}=-x \hat{i}+2 y \hat{j}-z \hat{k} \\ & \vec{a}=\frac{d \vec{v}}{d t}=-\frac{d x}{d t} \hat{i}+2 \frac{d y}{d t} \hat{j}-\frac{d z}{d t} \hat{k} \\ & =-(-x) \hat{i}+2(2 y) \hat{j}-(-z) \hat{k}=x \hat{i}+4 y \hat{j}+z \hat{k} \\ & \therefore \vec{a} \text { at }(1,2,4)=\hat{i}+8 \hat{j}+4 \hat{k} \\ & \therefore|\vec{a}|=9 \mathrm{~m} / \mathrm{s}^2\end{aligned}$

Question 10: A paratrooper jumps from an aeroplane and opens a paracliute after $2 s$ of free fall and starts deaccelerating with $3 \mathrm{~m} / \mathrm{s}^2$. At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is $5 \mathrm{~m} / \mathrm{s}$. The initial height of the airplane is... m. $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$

(1) 92.5

(2) 20

(3) 82.5

(4) 62.5

Correct Answer: (1)

Solution:

$A B=h_1, B C=h_2, C D=10 \mathrm{~m}$

$\begin{aligned}
& v_1=g t=20 \mathrm{~m} / \mathrm{s} \\
& h_1=\frac{1}{2} g t^2=20 \mathrm{~m} \\
& 5^2-v_1^2=2 \mathrm{as}=2(-3)\left(h_2\right) \\
& h_2=62.5 \mathrm{~m}
\end{aligned}$

Total height

$20+62.5+10=92.5$

Question 11: Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. $T_1$ and $T_2$ are the total flying times of first and second ball, respectively, then the ratio of $T_1$ and $T_2$ is :

(1) $2 \sqrt{2}: 1$

(2) $2: 1$

(3) $\sqrt{2}: 1$

(4) $4: 1$

Correct Answer: (1)

Solution:

Given, $\left(\mathrm{H}_{\max }\right)_1=8 \times\left(\mathrm{H}_{\max }\right)_2$

$
\begin{aligned}
& \frac{\mathrm{u}^2 \sin ^2 \theta_1}{2 \mathrm{~g}}=8 \times \frac{\mathrm{u}^2 \sin ^2 \theta_2}{2 \mathrm{~g}} \\
& \Rightarrow \sin \theta_1=2 \sqrt{2} \sin \theta_2 \\
& \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\frac{2 \mathrm{u} \sin \theta_1 / \mathrm{g}}{2 \mathrm{u} \sin \theta_2 / \mathrm{g}}=\frac{\sin \theta_1}{\sin \theta_2}=2 \sqrt{2}
\end{aligned}
$

Question 12: A helicopter flying horizontally with a speed of $360 \mathrm{~km} / \mathrm{h}$ at an altitude of 2 km , drops an object at an instant. The object hits the ground at a point O , 20 s after it is dropped. Displacement of ' O ' from the position of helicopter where the object was released is :

(use acceleration due to gravity $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ and neglect air resistance)

(1) $2 \sqrt{5} \mathrm{~km}$

(2) 4 km

(3) 7.2 km

(4) $2 \sqrt{2} \mathrm{~km}$

Correct Answer: (4)

Solution:

$\begin{aligned} & \mathbf{V}=360 \frac{\mathrm{~km}}{\mathrm{~h}}=360 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=100 \mathrm{~m} / \mathrm{sec} \\ & \mathbf{h}=2 \mathbf{k m} \\ & \mathrm{~T}=20 \mathrm{sec} \\ & \text { length 'Ho' }=\sqrt{\mathrm{H}^2+\mathbf{R}^2}=200 \sqrt{2}=2 \sqrt{2} \mathrm{~km}\end{aligned}$

Question 13: A particle moves along the $x$-axis and has its displacement x varying with time t according to the equation

$
\mathrm{x}=\mathrm{c}_0\left(\mathrm{t}^2-2\right)+\mathrm{c}(\mathrm{t}-2)^2
$

where $c_0$ and $c$ are constants of appropriate dimensions. Then, which of the following statements is correct?

(1) the acceleration of the particle is $2 \mathrm{c}_0$

(2) the acceleration of the particle is 2 c

(3) the initial velocity of the particle is 4 c

(4) the acceleration of the particle is $2\left(c+c_0\right)$

Correct Answer: (4)

Solution:

$\begin{aligned} & \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{tC}_0+2 \mathrm{C}(\mathrm{t}-2) \\ & \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{C}_0+2 \mathrm{C}\end{aligned}$

Question 14: 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 $\frac{n u^2}{25 g}$, where value of $n$ is :

(Given ' $g$ ' is the acceleration due to gravity).

(1) 6

(2) 18

(3) 12

(4) 24

Correct Answer: (4)

Solution:

$\begin{aligned} & R=3 H \\ & \frac{u^2 \sin 2 \theta}{g}=\frac{3 u^2 \sin ^2 \theta}{2 g} \\ & 2 \sin \theta \cos \theta=\frac{3}{2} \sin \theta \sin \theta \\ & \frac{4}{3}=\tan \theta \\ & \theta=53^{\circ} \\ & R=\frac{u^2}{g} \times 2 \times \frac{4}{5} \times \frac{3}{5}=\frac{24}{25} \frac{u^2}{g}\end{aligned}$

Question 15: A person travelling on a straight line moves with a uniform velocity $\mathrm{v}_1$ for a distance x and with a uniform velocity $\mathrm{v}_2$ for the next $\frac{3}{2} \mathrm{x}$ distance. The average velocity in this motion is $\frac{50}{7} \mathrm{~m} / \mathrm{s}$. If $\mathrm{v}_1$ is $5 \mathrm{~m} / \mathrm{s}$ then $\mathrm{v}_2=$ _____ $\mathrm{m} / \mathrm{s}$.

(1) 10

(2) 12

(3) 11

(4) 15

Correct Answer: (1)

Solution:

$
\begin{aligned}
& \mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{x}_1+\mathrm{x}_2}{\mathrm{t}_1+\mathrm{t}_2} \\
& \Rightarrow \frac{50}{7}=\frac{\mathrm{x}+\frac{3 \mathrm{x}}{2}}{\frac{\mathrm{x}}{5}+\frac{3 \mathrm{x}}{2 \mathrm{v}_2}} \\
& \Rightarrow \frac{50}{7}=\frac{5 / 2}{\frac{1}{5}+\frac{3}{2 \mathrm{v}_2}} \\
& \Rightarrow \frac{1}{5}+\frac{3}{2 \mathrm{v}_2}=\frac{7}{20} \\
& \Rightarrow \frac{3}{2 \mathrm{v}_2}=\frac{7}{20}-\frac{1}{5}=\frac{7-4}{20} \\
& \Rightarrow \frac{3}{2 \mathrm{v}_2}=\frac{3}{20} \\
& \Rightarrow \mathrm{v}_2=10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$

Also Read: JEE Mains 2027: Physics Set of 5 Sample Paper with Solution

Important Kinematics Topics for JEE Main 2027

Refer to the topics given below from which most JEE Main kinematics questions are asked. These topics are most important and can be asked in JEE Main 2027.

1. Distance and Displacement

2. Speed and Velocity

3. Motion: Uniform and Non-uniform

4. Equations of Motion

5. Graphs of Motion

6. Relative Motion

7. Projectile Motion

8. River Boat Problems

9. Rain Man Problems

Best Books for Kinematics JEE Main Questions

Refer to the books given below for solving Kinematics JEE questions. While preparing for IIT JEE, it is very important to follow the Best Books for JEE Main 2027 because they help properly cover all the concepts.