Projectiles A and B are thrown at angles of 45 and 60 with vertical respectively from top of a 400 ~m high tower. If…

Solution & Explanation

Core Logic

Projectile Motion from a Tower diagram for Q50 - JEE Main 2024 Evening

For two projectiles launched from the same height to have the same time of flight ( $T$ $T$), their vertical components of velocity must be equal. Since the angles given are with the vertical, the vertical components are $vA (45°)$ $v_A \cos(45^{\circ})$ and $vB (60°)$ $v_B \cos(60^{\circ})$. If $TA = TB$ $T_A = T_B$, then $vAy = vBy$ $v_{Ay} = v_{By}$.

vA (45°) = vB (60°)

vA ( 1\sqrt(2)) = vB ((1)/(2))

(vA)/(vB) = \sqrt(2)2 = 1\sqrt(2)

Step 1: Check Inconsistency

For the ranges to be the same while having the same time of flight, their horizontal components of velocity must also be equal: $vAx = vBx$ $v_{Ax} = v_{Bx}$.

vA (45°) = vB (60°)

vA ( 1\sqrt(2)) = vB ( \sqrt(3)2)

(vA)/(vB) = \sqrt((3)/(2))

This yields a contradiction. It is impossible for both the ranges and the times of flight to be simultaneously equal for different angles of projection from a tower.

Step 2: Conclusion

The question contains inconsistent data and is technically a Bonus question. However, if one arbitrarily equates only the time of flight (or if NTA intended a different scenario), the ratio $(vA)/(vB) = 1\sqrt(2)$ $\frac{v_A}{v_B} = \frac{1}{\sqrt{2}}$ matches option (4), which was the officially provided key before corrections.

Pattern Recognition

Be wary of over-constrained physics problems. If a question specifies both range AND time of flight are identical for two different angles, check if the math yields a contradiction. NTA often accepts the result of one partial constraint By NTA 4 BY Rankbit (Bonus).

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

Core Logic

Step 1: Check Inconsistency

Step 2: Conclusion

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

Core Logic

Step 1: Check Inconsistency

Step 2: Conclusion

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_30_january_evening

Select Your Target Exam

Full Mock Test Hub

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