JEE PYQ: Motion in a Plane - Question ID 4dd866733ec2 (JEE Main 2022)

Core Formula & Concept:

In projectile motion, the motion of an object launched at an angle can be resolved into two independent components:

• Horizontal motion: Uniform motion with constant velocity since no acceleration acts horizontally (ignoring air resistance). The horizontal velocity component is: $v_{x} = v_{0} \cos \alpha$ where $v_{0}$ is the initial velocity and $\alpha$ is the launch angle.
• Vertical motion: Uniformly accelerated motion under gravity. The vertical velocity component at any time $t$ is: $v_{y} = v_{0} \sin \alpha - g t$ where $g = 10 \, \text{ms}^{-2}$ is the acceleration due to gravity.

The inclination $\beta$ of the projectile with the horizontal at any time $t$ is the angle whose tangent is the ratio of the vertical velocity component to the horizontal velocity component: $\tan \beta = \frac{v_{y}}{v_{x}}$ This is the key formula we will use to find $\tan \beta$ after 10 seconds.

--- Step-by-Step Derivation:

Step 1: Write down initial velocity components
Initial velocity $v_{0} = 20 \, \text{ms}^{-1}$, angle $\alpha$
Horizontal component: $v_{x} = v_{0} \cos \alpha = 20 \cos \alpha$
Vertical component at $t = 0$: $v_{y0} = v_{0} \sin \alpha = 20 \sin \alpha$

Step 2: Find vertical velocity at $t = 10$ s
Using $v_{y} = v_{y0} - g t$, and $g = 10 \, \text{ms}^{-2}$
$v_{y} = 20 \sin \alpha - 10 \times 10 = 20 \sin \alpha - 100$

Step 3: Horizontal velocity remains constant
$v_{x} = 20 \cos \alpha$ (unchanged)

Step 4: Compute $\tan \beta$
$\tan \beta = \frac{v_{y}}{v_{x}} = \frac{20 \sin \alpha - 100}{20 \cos \alpha}$
Simplify numerator: $20 \sin \alpha - 100 = 20 (\sin \alpha - 5)$
So, $\tan \beta = \frac{20 (\sin \alpha - 5)}{20 \cos \alpha} = \frac{\sin \alpha - 5}{\cos \alpha}$ Split the fraction: $\tan \beta = \frac{\sin \alpha}{\cos \alpha} - \frac{5}{\cos \alpha} = \tan \alpha - 5 \sec \alpha$ since $\sec \alpha = \frac{1}{\cos \alpha}$

Step 5: Match with given options
We have derived: $\tan \beta = \tan \alpha - 5 \sec \alpha$ This matches Option B.

--- Common Traps & Exam Tip:

Trap 1: Forgetting that horizontal velocity is constant
Students often mistakenly apply acceleration to the horizontal component. Remember: $v_{x}$ does not change.

Trap 2: Sign error in vertical velocity
The vertical velocity decreases due to gravity, so $v_{y} = v_{0} \sin \alpha - g t$. A common mistake is using $+g t$, which would imply upward acceleration.

Trap 3: Misinterpreting $\tan \beta$
$\tan \beta$ is the ratio of current vertical velocity to horizontal velocity, not initial velocity components.

Exam Tip: Always resolve motion into horizontal and vertical components. Use $\tan \theta = \frac{v_{y}}{v_{x}}$ for inclination at any instant. Double-check signs and units.