A ball rolls off the top of a stairway with horizontal velocity u. The steps are 0.1 ~m high and 0.1 ~m wide. The…

Solution & Explanation

Related Formula

For a projectile fired horizontally from height $h$ $h$ with speed $u$ $u$:

Horizontal Range R = u \cdot t

Vertical Displacement h = (1)/(2) g t²

Core Logic

To just clear step 4 and land on the tread of step 5, the flight trajectory must pass beyond the outer corner boundary vertex of the $4th$ $4^{\text{th}}$ step.

Therefore, net horizontal distance required to cross 4 complete steps is:

R = 4 \times 0.1 ~m = 0.4 ~m

Similarly, net vertical fall matching the top of the $4th$ $4^{\text{th}}$ step level is:

h = 4 \times 0.1 ~m = 0.4 ~m

Trajectory diagram of a ball clearing step corners on a staircase grid for Q52

Step 1: Determine Time of Flight

Using the vertical kinematic equation:

0.4 = (1)/(2) \times 10 \times t² 0.4 = 5 t²

t² = (0.4)/(5) = 0.08 ~s²

Step 2: Determine Velocity

Using the horizontal path equation:

R = u \cdot t R² = u² \cdot t²

Substituting range and time values:

(0.4)² = u² \times 0.08

0.16 = u² \times 0.08

u² = (0.16)/(0.08) = 2 u = \sqrt(2) ~m/s

Step 3: Extract x

Matching the parameter form $u = \sqrt(x)$ $u = \sqrt{x}$, we find:

$x = 2$ $x = 2$

Pattern Recognition

To clear the $n$ $n$-th step, the projectile must safely pass the outer point of step $(n-1)$ $(n-1)$. Treat the geometric coordinates of corners as bounding constraints ( $x = (n-1)w$ $x = (n-1)w$, $y = (n-1)h$ $y = (n-1)h$) to configure kinematics instantaneously.

Chapter Mix

Class 11 Physics: Motion in a Plane

More Motion in a Plane 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: Determine Time of Flight

Step 2: Determine Velocity

Step 3: Extract x

Pattern Recognition

Chapter Mix

More Motion in a Plane 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: Determine Time of Flight

Step 2: Determine Velocity

Step 3: Extract x

Pattern Recognition

Chapter Mix

More Motion in a Plane 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 Motion in a Plane Questions — jee_main_2024_29_jan_morning

Select Your Target Exam

Full Mock Test Hub

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