Kinematics Previous Year Questions — JEE Main Physics
38 past-year Kinematics questions from JEE Main (Physics) — page 7.
Q39 (2026)
A river of width $200 \text{ m}$ is flowing from west to east with a speed of $18 \text{ km/h}$. A boat, moving with speed of $36 \text{ km/h}$ in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ________ and ________ respectively.
- $20 \text{ s and } 100 \text{ m}$
- $40 \text{ s and } 0 \text{ m}$
- $40 \text{ s and } 200 \text{ m}$
- $40 \text{ s and } 100 \text{ m}$
### Related Formula
$$t_{\text{min}} = \frac{d}{v_{br}}$$
$$\text{Drift (displacement along bank)} = v_r \times t_{\text{total}}$$
### Core Logic
First, convert speeds to m/s.
River speed $v_R = 18 \times \frac{5}{18} = 5 \text{ m/s}$.
Boat speed in still water $v_{BR} = 36 \times \frac{5}{18} = 10 \text{ m/s}$.
River width $d = 200 \text{ m}$.
{{SOL_IMG1}}
For minimum time across the river, the boat must head exactly perpendicular to the river flow.
### Step 1: Calculating Minimum Time
Time taken for one crossing (bank to bank):
$$t_{\text{one-way}} = \frac{d}{v_{BR}} = \frac{200}{10} = 20 \text{ s}$$
Since it is a round trip (bank to bank and back), total minimum time is:
$$t_{\text{total}} = 2 \times 20 = 40 \text{ s}$$
### Step 2: Calculating Displacement along Bank
Displacement along the river bank (drift) happens exclusively due to the river's flow during this total time:
$$\text{Displacement} = v_R \times t_{\text{total}} = 5 \times 40 = 200 \text{ m}$$
### Pattern Recognition
Minimum time always ignores drift and maxes out the perpendicular velocity vector. Drift simply becomes $V_{\text{river}} \times T_{\text{total}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
Q32 (2026)
A projectile is thrown upward at an angle $60^{circ}$ with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is $45^{circ}$ with the horizontal. The initial speed of the projectile is \_\_\_\_ m/s.
- $40\sqrt{2}$
- 40
- $20\sqrt{3}$
- $20\sqrt{2}$
### Related Formula
$$u_x = u \cos\theta = v \cos\phi$$
### Core Logic
{{SOL_IMG1}}
Since horizontal component of velocity remains constant:
$$u \cos 60^{\circ} = 20 \cos 45^{\circ}$$
$$\frac{u}{2} = \frac{20}{\sqrt{2}}$$
$$u = \frac{40}{\sqrt{2}} = 20\sqrt{2} \text{ m/s}$$
### Pattern Recognition
Sees: Projectile motion speed at intermediate angle.
Shortcut: Equate horizontal velocity components before and during flight.
Check: Matches option (4). ✓
### Chapter Mix
Class 11 Physics: Motion in a Plane
Q39 (2026)
An object is projected with kinetic energy K from a point A at an angle $60^{\circ}$ with the horizontal. The ratio of the difference in kinetic energies points B and C to that at point A (see figure), in the absence of air friction is :
{{IMG1}}
- $1:2$
- $2:3$
- $1:4$
- $3:4$
### Related Formula
$$K = \frac{1}{2}mu^{2}$$
$$K_{x} = \frac{1}{2}m(u \cos \theta)^{2} = K \cos^{2} \theta$$
### Core Logic
For a projectile with no air resistance, horizontal velocity remains constant. At the highest point (B), the vertical velocity is zero, so the kinetic energy is entirely due to the horizontal velocity. At point C, it hits the ground with the same speed as projected, restoring full kinetic energy.
### Step 1: Find KE at distinct points
$$(\text{KE})_{A} = K = \frac{1}{2}mu^{2}$$
At maximum height (B), velocity is $v_{B} = u \cos 60^{\circ} = \frac{u}{2}$.
$$(\text{KE})_{B} = \frac{1}{2}m\left(\frac{u}{2}\right)^{2} = \frac{1}{4}\left(\frac{1}{2}mu^{2}\right) = \frac{K}{4}$$
At landing point (C), speed is identical to initial speed $u$.
$$(\text{KE})_{C} = K$$
### Step 2: Calculate Ratio
The required ratio is between the difference in kinetic energies at points C and B, to that at point A:
$$\text{Ratio} = \frac{K_{C} - K_{B}}{K_{A}}$$
$$\text{Ratio} = \frac{K - \frac{K}{4}}{K} = \frac{\frac{3K}{4}}{K} = \frac{3}{4}$$
### Pattern Recognition
Sees: "KE at highest point" → Substitute $K \cos^2 \theta$. For $\theta = 60^{\circ}$, $\cos 60^{\circ} = 1/2$, so $K_B = K/4$. The difference with ground is $3K/4$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane
Q30 (2026)
A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with $3 \, \mathrm{m/s}^{2}$ . At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the aeroplane is ____ m.
(g = 10 m/s $^{2}$ )
- $62.5$
- $92.5$
- $20$
- $82.5$
### Related Formula
$$s = ut + \frac{1}{2}at^2$$
$v = u + at$
$v^2 = u^2 + 2as$
### Core Logic
The motion is divided into three distinct vertical sections:
{{SOL_IMG1}}
{{SOL_IMG2}}
1) Free fall phase ($A \to B$)
2) Deceleration phase with parachute ($B \to C$)
3) Final descent phase to the ground ($C \to D$)
### Step 1: Free Fall (A to B)
Initial velocity $u = 0$, $t = 2 \, \mathrm{s}$, $g = 10 \, \mathrm{m/s^2}$.
$$x_1 = \frac{1}{2} \times 10 \times 2^2 = 20 \, \mathrm{m}$$
Velocity at B:
$$V_B = 0 + 10 \times 2 = 20 \, \mathrm{m/s}$$
### Step 2: Deceleration Phase (B to C)
Starts with $V_B = 20 \, \mathrm{m/s}$, decelerates at $3 \, \mathrm{m/s^2}$ ($a = -3 \, \mathrm{m/s^2}$), and reaches $V_C = 5 \, \mathrm{m/s}$.
$$V_C^2 = V_B^2 + 2a x_2$$
$$5^2 = 20^2 - 2(3)x_2$$
$25 = 400 - 6x_2$
$$6x_2 = 375 \implies x_2 = \frac{375}{6} = 62.5 \, \mathrm{m}$$
### Step 3: Final Phase and Total Height
The last phase $x_3$ is given as $10 \, \mathrm{m}$ above the ground.
$$H = x_1 + x_2 + x_3$$
$$H = 20 + 62.5 + 10 = 92.5 \, \mathrm{m}$$
### Pattern Recognition
Split multi-stage motion problems into clear intervals. The final velocity of interval N becomes the initial velocity of interval N+1. Calculate displacement for each interval independently and sum them up.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
Q43 (2026)
A boy thrown a ball into air at $45^{\circ}$ from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is ____ m.
($g = 10 \text{ m/s}^{2}$)
- $20$
- $10$
- $25$
- $15$
### Related Formula
$$t_{\text{max_height}} = \frac{u_y}{g}$$
$$y = u_y t - \frac{1}{2} g t^2$$
### Core Logic
{{SOL_IMG1}}
The time taken to reach the maximum height is given as $2 \text{ s}$.
This means:
$$t_{\text{peak}} = \frac{u_y}{g} = 2$$
$$u_y = 2 \times 10 = 20 \text{ m/s}$$
### Step 1: Calculate Building Height
The ball lands on the roof at $t = 3 \text{ s}$. The height $H$ at this instant is the vertical displacement $y$.
Using the kinematic equation for vertical motion:
$$H = u_y t - \frac{1}{2} g t^2$$
$$H = (20)(3) - \frac{1}{2}(10)(3)^2$$
$H = 60 - 5(9)$
$$H = 60 - 45 = 15 \text{ m}$$
### Pattern Recognition
The launch angle ($45^{\circ}$) is distractor data! Maximum height timing strictly dictates initial vertical velocity, which is all you need for the vertical height calculation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane
