JEE PYQ: Motion in a Plane - Question ID c71d14988d61 (JEE Main 2022)

Core Formula & Concept:

In projectile motion, the velocity of the projectile can be resolved into horizontal ($v_x$) and vertical ($v_y$) components. The key formulas used are:

• Initial velocity components: $v_x = v_0 \cos \theta$ $v_y = v_0 \sin \theta - gt$
• Inclination of velocity vector with horizontal: The angle $\phi$ that the velocity vector makes with the horizontal at any time $t$ is given by: $\tan \phi = \frac{v_y}{v_x}$
• Condition given in the problem: At time $t$, the inclination $\phi$ becomes zero, meaning the projectile is moving horizontally at that instant. This implies: $v_y = 0$
• Horizontal range $R$: The horizontal range of a projectile is given by: $R = \frac{v_0^2 \sin 2\theta}{g}$

Using these concepts, we will derive the value of $\theta$ in terms of $R$ and $t$.

--- Step-by-Step Derivation:

Step 1: Use the condition for zero inclination at time $t$

At time $t$, the vertical component of velocity becomes zero: $v_y = v_0 \sin \theta - gt = 0$

Solving for $t$: $v_0 \sin \theta = gt$ $\sin \theta = \frac{gt}{v_0}$

Given $v_0 = 25$ m/s and $g = 10$ m/s²: $\sin \theta = \frac{10t}{25} = \frac{2t}{5}$

Step 2: Express the horizontal range $R$ in terms of $\theta$

The horizontal range is: $R = \frac{v_0^2 \sin 2\theta}{g} = \frac{25^2 \cdot \sin 2\theta}{10} = \frac{625 \sin 2\theta}{10} = 62.5 \sin 2\theta$

Using the double-angle identity: $\sin 2\theta = 2 \sin \theta \cos \theta$

Substituting $\sin \theta = \frac{2t}{5}$ from Step 1: $\sin 2\theta = 2 \cdot \frac{2t}{5} \cdot \cos \theta = \frac{4t}{5} \cos \theta$

Thus: $R = 62.5 \cdot \frac{4t}{5} \cos \theta = 50 t \cos \theta$

Solving for $\cos \theta$: $\cos \theta = \frac{R}{50 t}$

Step 3: Relate $\sin \theta$ and $\cos \theta$ to find $\theta$

We have: $\sin \theta = \frac{2t}{5}$ $\cos \theta = \frac{R}{50 t}$

Using the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$ $\left(\frac{2t}{5}\right)^2 + \left(\frac{R}{50 t}\right)^2 = 1$ $\frac{4t^2}{25} + \frac{R^2}{2500 t^2} = 1$

However, instead of solving this equation, we can directly find $\cot \theta$: $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{R}{50 t}}{\frac{2t}{5}} = \frac{R}{50 t} \cdot \frac{5}{2t} = \frac{R}{20 t^2}$

Thus: $\theta = \cot^{-1} \left( \frac{R}{20 t^2} \right)$

Step 4: Match with the given options

The derived expression for $\theta$ is: $\theta = \cot^{-1} \left( \frac{R}{20 t^2} \right)$

This matches Option D.

--- Common Traps & Exam Tip:

1. Misinterpreting the condition for zero inclination: Many students confuse the condition for zero inclination with the time of flight. The time of flight is when the projectile returns to the ground ($y = 0$), whereas zero inclination occurs when the vertical velocity becomes zero ($v_y = 0$).

2. Incorrect use of range formula: Students often forget to use the double-angle identity for $\sin 2\theta$ or make algebraic mistakes while substituting values. Always verify the substitution of $\sin \theta$ and $\cos \theta$ carefully.

3. Overcomplicating the trigonometric relations: Instead of solving for $\theta$ directly, it is often easier to find $\cot \theta$ or $\tan \theta$ first, as done in this solution. This avoids unnecessary quadratic equations.

Exam Tip: When dealing with projectile motion problems involving angles and ranges, always:

• Resolve the velocity into components.
• Use the condition given (here, $v_y = 0$ at time $t$) to find a relation involving $\theta$.
• Express the range $R$ in terms of $\theta$ and substitute known values.
• Look for trigonometric identities to simplify the expression.