An object is projected with kinetic energy K from a point A at an angle 60 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 :

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. At maximum height (B), velocity is vB = u 60 = (u/2). At landing point (C), speed is identical to initial speed u. The required ratio is between the difference in kinetic energies at points C and B, to that at point A: Sees: "KE at highest point" → Substitute K 2 theta. For theta = 60, 60 = 1/2, so KB = K/4. The difference with ground is 3K/4. Class 11 Physics: Motion in a Plane

An object is projected with kinetic energy K from a point A at an angle 60 with the horizontal. The ratio of the difference in kinetic energies points B and C to that at point A (see figure), in the a

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. At maximum height (B), velocity is vB = u 60 = (u/2). At landing point (C), speed is identical to initial speed u. The required ratio is between the difference in kinetic energies at points C and B, to

An object is projected with kinetic energy K from a point A at an angle 60 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 :

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. At maximum height (B), velocity is vB = u 60 = (u/2). At landing point (C), speed is identical to initial speed u. The required ratio is between the difference in kinetic energies at points C and B, to that at point A: Sees: "KE at highest point" → Substitute K 2 theta. For theta = 60, 60 = 1/2, so KB = K/4. The difference with ground is 3K/4. Class 11 Physics: Motion in a Plane

An object is projected with kinetic energy K from a point A at an angle 60 with the horizontal. The ratio of the difference in kinetic energies points B and C to that at point A (see figure), in the a

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. At maximum height (B), velocity is vB = u 60 = (u/2). At landing point (C), speed is identical to initial speed u. The required ratio is between the difference in kinetic energies at points C and B, to

An object is projected with kinetic energy K from a point A at an angle 60 with the horizontal. The ratio of the…

Q8 jee_main_2025_24_jan_evening Velocity and Acceleration

The position vector of a moving body at any instant of time is given as

r = (5t² i - 5t j)m

. The magnitude and direction of velocity at

t = 2s

is,

A. $5\sqrt(15)$ $5\sqrt{15}$ m/s, making an angle of $⁻¹4$ $\tan^{-1}4$ with -ve Y axis
B. $5\sqrt(15)$ $5\sqrt{15}$ m/s, making an angle of $⁻¹4$ $\tan^{-1}4$ with +ve X axis
C. $5\sqrt(17)$ $5\sqrt{17}$ m/s, making an angle of $⁻¹4$ $\tan^{-1}4$ with -ve Y axis
D. $5\sqrt(17)$ $5\sqrt{17}$ m/s, making an angle of $⁻¹4$ $\tan^{-1}4$ with +ve X axis

Solution

Related Formula

v = d rdt

Core Logic

Given position vector:

r = 5t² i - 5t j

Differentiating with respect to $t$ $t$:

v = 10t i - 5 j

At $t = 2 s$ $t = 2\ \mathrm{s}$:

v = 20 i - 5 j

Magnitude of velocity:

v = \sqrt((20)² + (-5)²) = \sqrt(400 + 25) = \sqrt(425) = 5\sqrt(17) m/s

Direction analysis:

Velocity vector components angle calculation Q8

vₓ = 20

along

+x

axis,

vy = -5

along

-y

axis. Let

θ

be the angle made with the negative

Y

-axis:

θ = |(vₓ)/(vy)| = (20)/(5) = 4 θ = ⁻¹4

Pattern Recognition

Always read the reference axis carefully in direction questions. Here, the angle is measured from the negative $Y$ $Y$-axis, making $θ = vₓ / |vy|$ $\tan\theta = v_x / |v_y|$.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q9 jee_main_2025_28_jan_evening Velocity Time Graphs

The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between

t = 0

to

t = 4s

?

Velocity Time Graphs diagram for Q9 - JEE Main 2025 Evening — The velocity-time profile tracking motion across consecutive geometric shapes up to 4 seconds.

A. $30m$ $30\mathrm{m}$
B. $10m$ $10\mathrm{m}$
C. $13m$ $13\mathrm{m}$
D. $11m$ $11\mathrm{m}$

Solution

Related Formula

The distance traveled by an object equals the total area enclosed under its velocity-time ( $v-t$ $v-t$) plot along the time axis, treating all regional boundaries as strictly positive metrics:

Distance = \int |v| dt

Core Logic

From the geometric grid profile between $t = 0$ $t = 0$ and $t = 4 s$ $t = 4\text{ s}$:

First Region ( $t=0$ $t=0$ to $t=2 s$ $t=2\text{ s}$): Forms a \right-angled \triangle with base $= 2 s$ $= 2\text{ s}$ and peak height $= 10 ms⁻¹$ $= 10\text{ ms}^{-1}$.

Area₁ = (1)/(2) \times 2 \times 10 = 10 m

Second Region ( $t=2$ $t=2$ to $t=4 s$ $t=4\text{ s}$): Forms a standard rectangle with width $= (4 - 2) = 2 s$ $= (4 - 2) = 2\text{ s}$ and height $= 10 ms⁻¹$ $= 10\text{ ms}^{-1}$.

Area₂ = 2 \times 10 = 20 m

Summing the areas together to extract total displacement path:

Total Distance = 10 + 20 = 30 m

Pattern Recognition

Always differentiate between distance and displacement on graph tracks. Displacement treats components below the axis as negative fields, while distance calculates absolute geometric magnitudes without direction.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Q jee_main_2025_29_jan_morning Projectile Motion

Two projectiles are fired with same initial speed from same point on ground at angles of

(45° - α)

and

(45° + α)

, respectively, with the horizontal direction. The ratio of their maximum heights attained is :

A. $(1 - α)/(1 + α)$ $\frac{1 - \tan\alpha}{1 + \tan\alpha}$
B. $(1 + α)/(1 - α)$
C. $(1 - 2α)/(1 + 2α)$
D. $(1 + 2α)/(1 - 2α)$

Solution

Related Formula

H = (u² ² θ)/(2g)

Core Logic

Let $θ₁ = 45° - α$ $\theta_1 = 45^{\circ} - \alpha$ and $θ₂ = 45° + α$ $\theta_2 = 45^{\circ} + \alpha$. The ratio of maximum heights is:

(H₁)/(H₂) = ²(45° - α) ²(45° + α)

Step 1: Simplify Trigonometric Terms

(45° \pm α) = 1\sqrt(2) α \pm 1\sqrt(2) α

(H₁)/(H₂) = (( α - α)²)/(( α + α)²) = ( ²α + ²α - 2 α α)/( ²α + ²α + 2 α α) = (1 - 2α)/(1 + 2α)

Pattern Recognition

For complementary angles shifted symmetric to $45°$ $45^{\circ}$, ratios simplify cleanly via $2α$ $\sin 2\alpha$ identities.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q jee_main_2025_29_jan_morning Relative Motion

The maximum speed of a boat in still water is

27 km/h

. Now this boat is moving downstream in a river flowing at

9 km/h

. A man in the boat throws a ball vertically upwards with speed of

10 m/s

. Range of the ball as observed by an observer at rest on the river bank, is ________ cm. (Take

g = 10 m/s²

)

Numerical Answer. Answer: 2000 to 2000

Solution

Related Formula

T = (2uy)/(g), R = vₓ \cdot T

Core Logic

The total horizontal velocity components combine due to downstream motion addition

vₓ = 27 + 9 = 36 km/h = 36 \cdot (5)/(18) = 10 m/s

The time of flight for the vertical launch tracking stands at :

T = (2 \cdot 10)/(10) = 2 s

Horizontal range measured along the river bank frame is :

Range = vₓ \cdot T = 10 \cdot 2 = 20 m = 2000 cm

Chapter Mix

Class 11 Physics: Motion in a Straight Line Class 11 Physics: Motion in a Plane

Q47 jee_main_2024_01_february_morning Projectile Motion

A particle moving in a circle of radius

R

with uniform speed takes time

T

to complete one revolution. If this particle is projected with the same speed at an angle

θ

to the horizontal, the maximum height attained by it is equal to

4R

. The angle of projection

θ

is then given by:

A. $⁻¹[ 2gT²pi²R](1)/(2)$
B. $⁻¹[ π²R2gT²](1)/(2)$
C. $⁻¹[ 2gT²pi²R](1)/(2)$
D. $⁻¹[(π R)/(2gT²)](1)/(2)$

Solution

Related Formula

Uniform circular velocity:

v = (2π R)/(T)

Maximum height of a projectile:

H = (v² ²θ)/(2g)

Core Logic

Given that the projectile max height matches $H = 4R$ $H = 4R$:

4R = (v² ²θ)/(2g)

Substitute the value of $v$ $v$ from the circular motion loop:

4R = (((2π R)/(T))² ²θ)/(2g)

4R = (4π² R² ²θ)/(2g T²)

Step 1: Isolate Angular Components

Cancel out $4R$ $4R$ from both sides:

1 = (π² R ²θ)/(2g T²)

²θ = (2g T²)/(π² R)

θ = ((2g T²)/(π² R))(1)/(2) θ = ⁻¹[(2gT²)/(π² R)](1)/(2)

Pattern Recognition

Connect circular metrics directly to projectile parameters via velocity matching. Keeping terms unsimplified makes it easy to cancel common elements later.

Chapter Mix

Class 11 Physics: Motion in a Plane

Solution & Explanation

Related Formula

Core Logic

Step 1: Find KE at distinct points

Step 2: Calculate Ratio

Pattern Recognition

Chapter Mix

Reference Study Guides

More Motion in a Plane Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Trigonometric Terms

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Isolate Angular Components

Pattern Recognition

Chapter Mix

More Motion in a Plane Questions — jee_main_2026_23_january_morning

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