A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 , m/s2 . 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 )

v = u + at v2 = u2 + 2as The motion is divided into three distinct vertical sections: 1) Free fall phase (A B) 2) Deceleration phase with parachute (B C) 3) Final descent phase to the ground (C D) Initial velocity u = 0, t = 2 , s, g = 10 , m/s2. Velocity at B: Starts with VB = 20 , m/s, decelerates at 3 , m/s2 (a = -3 , m/s2), and reaches VC = 5 , m/s. 25 = 400 - 6x2 The last phase x3 is given as 10 , m above the ground. 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. Class 11 Physics: Kinematics

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 , m/s2 . At 10 m height from ground, while descending with the help of parachute, th

v = u + at v2 = u2 + 2as The motion is divided into three distinct vertical sections: 1) Free fall phase (A B) 2) Deceleration phase with parachute (B C) 3) Final descent phase to the ground (C D) Initial velocity u = 0, t = 2 , s, g = 10 , m/s2. Velocity at B: Starts with VB = 20 , m/s, decelerates at 3 , m/s2 (a = -3 , m/s2), and reaches VC = 5 , m/s. 25 = 400 - 6x2 The last phase x3 is given as 10 , m above the ground. Split multi-stage motion problems into clear intervals. The final velocity

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 , m/s2 . 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 )

v = u + at v2 = u2 + 2as The motion is divided into three distinct vertical sections: 1) Free fall phase (A B) 2) Deceleration phase with parachute (B C) 3) Final descent phase to the ground (C D) Initial velocity u = 0, t = 2 , s, g = 10 , m/s2. Velocity at B: Starts with VB = 20 , m/s, decelerates at 3 , m/s2 (a = -3 , m/s2), and reaches VC = 5 , m/s. 25 = 400 - 6x2 The last phase x3 is given as 10 , m above the ground. 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. Class 11 Physics: Kinematics

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 , m/s2 . At 10 m height from ground, while descending with the help of parachute, th

v = u + at v2 = u2 + 2as The motion is divided into three distinct vertical sections: 1) Free fall phase (A B) 2) Deceleration phase with parachute (B C) 3) Final descent phase to the ground (C D) Initial velocity u = 0, t = 2 , s, g = 10 , m/s2. Velocity at B: Starts with VB = 20 , m/s, decelerates at 3 , m/s2 (a = -3 , m/s2), and reaches VC = 5 , m/s. 25 = 400 - 6x2 The last phase x3 is given as 10 , m above the ground. Split multi-stage motion problems into clear intervals. The final velocity

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3…

Year	2026	2025	2024	Total
Questions	8	16	14	38

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

s = ut + (1)/(2)at²

$v = u + at$ $v = u + at$ $v² = u² + 2as$ $v^2 = u^2 + 2as$

Core Logic

The motion is divided into three distinct vertical sections:

Motion under Gravity diagram for Q30 - JEE Main 2026 Evening

Free fall phase ( $A \to B$ $A \to B$)
Deceleration phase with parachute ( $B \to C$ $B \to C$)
Final descent phase to the ground ( $C \to D$ $C \to D$)

Step 1: Free Fall (A to B)

Initial velocity $u = 0$ $u = 0$, $t = 2 s$ $t = 2 \, \mathrm{s}$, $g = 10 m/s²$ $g = 10 \, \mathrm{m/s^2}$.

x₁ = (1)/(2) \times 10 \times 2² = 20 m

Velocity at B:

VB = 0 + 10 \times 2 = 20 m/s

Step 2: Deceleration Phase (B to C)

Starts with $VB = 20 m/s$ $V_B = 20 \, \mathrm{m/s}$, decelerates at $3 m/s²$ $3 \, \mathrm{m/s^2}$ ( $a = -3 m/s²$ $a = -3 \, \mathrm{m/s^2}$), and reaches $VC = 5 m/s$ $V_C = 5 \, \mathrm{m/s}$.

VC² = VB² + 2a x₂

5² = 20² - 2(3)x₂

$25 = 400 - 6x₂$ $25 = 400 - 6x_2$

6x₂ = 375 x₂ = (375)/(6) = 62.5 m

Step 3: Final Phase and Total Height

The last phase $x₃$ $x_3$ is given as $10 m$ $10 \, \mathrm{m}$ above the ground.

H = x₁ + x₂ + x₃

H = 20 + 62.5 + 10 = 92.5 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.

Chapter Mix

Class 11 Physics: Kinematics

More Kinematics Previous-Year Questions — Page 8

Q44 jee_main_2024_31_jan_evening Vector Algebra

If two vectors

A

and

B

having equal magnitude

R

are inclined at an angle

θ

, then

A. $| A - B| = \sqrt(2)R ((θ)/(2))$
B. $| A + B| = 2R ((θ)/(2))$
C. $| A + B| = 2R ((θ)/(2))$
D. $| A - B| = 2R ((θ)/(2))$

Solution

Related Formula

The magnitude of the resultant vector is given by:

| Rᵣₑₛ| = \sqrt(A² + B² + 2AB θ)

Core Logic

Let $| A| = | B| = R$ $|\vec{A}| = |\vec{B}| = R$. Then for vector addition:

| A + B| = \sqrt(R² + R² + 2R² θ)

Step 1: Simplify Addition Form

| A + B| = \sqrt(2R² (1 + θ))

Using the trigonometric identity $1 + θ = 2 ² ((θ)/(2))$ :

| A + B| = \sqrt(2R² \times 2 ² ((θ)/(2))) = 2R ((θ)/(2))

Step 2: Cross-check Subtraction Form

For subtraction:

| A - B| = \sqrt(R² + R² - 2R² θ)

| A - B| = \sqrt(2R² (1 - θ)) = \sqrt(2R² \times 2 ² ((θ)/(2))) = 2R ((θ)/(2))

Checking options, only $| A + B| = 2R ((θ)/(2))$ is correctly paired in the choice list.

Pattern Recognition

Standard geometry shortcut: Addition of two equal vectors yields a cosine half-angle dependency. Subtraction yields a sine half-angle dependency. $(+ \to )$ $(+ \to \cos)$, $(- \to )$ $(- \to \sin)$.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q jee_main_2024_31_jan_morning Projectile Motion

A body starts falling freely from height

H

hits an inclined plane in its path at height

h

. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of

(H)/(h)

for which the body will take the maximum time to reach the ground is

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

t = \sqrt((2d)/(g))

Core Logic

Projectile Motion diagram for Q55 - JEE Main 2024 Morning

The body falls freely from height $H$ $H$ to height $h$ $h$. The distance fallen is $(H - h)$ $(H - h)$. Time taken to fall this distance:

t₁ = \sqrt((2(H - h))/(g))

After elastic impact, the vertical velocity becomes zero (the entire velocity is directed horizontally). From height $h$ $h$, it now acts as a horizontal projectile. The time taken to reach the ground vertically from height $h$ $h$ is:

t₂ = \sqrt((2h)/(g))

Total time of flight $T = t₁ + t₂$ $T = t_1 + t_2$:

T = \sqrt((2(H - h))/(g)) + \sqrt((2h)/(g))

Step 2: Maximizing Time

To find the maximum time $T$ $T$, differentiate $T$ $T$ with respect to $h$ $h$ and equate to zero:

(dT)/(dh) = \sqrt((2)/(g)) ( -12\sqrt(H - h) + 12\sqrt(h) ) = 0

12\sqrt(h) = 12\sqrt(H - h)

\sqrt(H - h) = \sqrt(h)

Squaring both sides: $H - h = h$ $H - h = h$ $H = 2h$ $H = 2h$

(H)/(h) = 2

Chapter Mix

Class 11 Physics: Kinematics

Q33 jee_main_2024_31_jan_morning Differentiation In Kinematics

The relation between time '

t

' and distance '

x

' is

t = α x² + β x

, where

α

and

β

are constants. The relation between acceleration (

a

) and velocity (

v

) is:

A. $a = -2α v³$ $a = -2\alpha v^3$
B. $a = -5α v⁵$ $a = -5\alpha v^5$
C. $a = -3α v²$ $a = -3\alpha v^2$
D. $a = -4α v⁴$ $a = -4\alpha v^4$

Solution

Related Formula

v = (dx)/(dt)

a = (dv)/(dt) = v(dv)/(dx)

Step 1: Differentiate with respect to time

Given the equation:

t = α x² + β x

Differentiating with respect to time $t$ $t$:

(dt)/(dt) = (d)/(dt)(α x² + β x)

1 = 2α x (dx)/(dt) + β (dx)/(dt)

1 = (2α x + β) v

v = (2α x + β)⁻¹

Step 2: Calculate Acceleration

Now, acceleration $a = (dv)/(dt)$ $a = \frac{dv}{dt}$. Differentiating $v$ $v$ with respect to time $t$ $t$:

a = (d)/(dt) [ (2α x + β)⁻¹ ]

a = -1(2α x + β)⁻² \cdot (d)/(dt)(2α x + β)

a = -(2α x + β)⁻² \cdot (2α) (dx)/(dt)

Substitute $v$ $v$ and $(2α x + β)⁻² = v²$ $(2\alpha x + \beta)^{-2} = v^2$:

a = -(v²) \cdot (2α) \cdot v

a = -2α v³

Pattern Recognition

Standard kinematic shortcut: Whenever $t = Ax² + Bx$ $t = Ax^2 + Bx$, $v = (2Ax+B)⁻¹$ $v = (2Ax+B)^{-1}$ and $a = -2A v³$ $a = -2A v^3$. Memorizing this directly saves derivation time during the exam.

Chapter Mix

Class 11 Physics: Kinematics

Solution & Explanation

Related Formula

Core Logic

Step 1: Free Fall (A to B)

Step 2: Deceleration Phase (B to C)

Step 3: Final Phase and Total Height

Pattern Recognition

Chapter Mix

Reference Study Guides

More Kinematics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Simplify Addition Form

Step 2: Cross-check Subtraction Form

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Maximizing Time

Chapter Mix

Solution

Related Formula

Step 1: Differentiate with respect to time

Step 2: Calculate Acceleration

Pattern Recognition

Chapter Mix

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Solution & Explanation

Related Formula

Core Logic

Step 1: Free Fall (A to B)

Step 2: Deceleration Phase (B to C)

Step 3: Final Phase and Total Height

Pattern Recognition

Chapter Mix

Reference Study Guides

More Kinematics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Simplify Addition Form

Step 2: Cross-check Subtraction Form

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Maximizing Time

Chapter Mix

Solution

Related Formula

Step 1: Differentiate with respect to time

Step 2: Calculate Acceleration

Pattern Recognition

Chapter Mix

More Kinematics Questions — jee_main_2026_23_january_evening

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Full Mock Test Hub

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