JEE PYQ: Motion in a Plane - Question ID 3861c096ae9a (JEE Main 2022)

Core Formula & Concept:

In projectile motion, a body is launched at an angle to the horizontal and moves under the influence of gravity. The motion can be resolved into two independent components:

• Horizontal motion: Uniform motion with constant velocity (no acceleration).
• Vertical motion: Accelerated motion under gravity ($g = 10 \text{ ms}^{-2}$ downward).

Key formulas used:

• Initial velocity components: $v_{0x} = v_0 \cos \theta, \quad v_{0y} = v_0 \sin \theta$ where $v_0$ is the initial speed and $\theta = 45^\circ$ is the projection angle.
• Velocity components at time $t$: $v_x(t) = v_{0x}, \quad v_y(t) = v_{0y} - gt$
• Magnitude of velocity at time $t$: $v(t) = \sqrt{v_x(t)^2 + v_y(t)^2}$
• Maximum height reached: $H = \frac{v_{0y}^2}{2g}$

Step-by-Step Derivation:

Step 1: Resolve initial velocity

Given $\theta = 45^\circ$, we have: $v_{0x} = v_0 \cos 45^\circ = \frac{v_0}{\sqrt{2}}, \quad v_{0y} = v_0 \sin 45^\circ = \frac{v_0}{\sqrt{2}}$

Step 2: Express velocity at $t = 2\text{ s}$

At $t = 2\text{ s}$, the horizontal velocity remains unchanged: $v_x(2) = v_{0x} = \frac{v_0}{\sqrt{2}}$ The vertical velocity decreases due to gravity: $v_y(2) = v_{0y} - g \cdot 2 = \frac{v_0}{\sqrt{2}} - 20$ The magnitude of velocity at $t = 2\text{ s}$ is given as $20 \text{ ms}^{-1}$: $v(2) = \sqrt{v_x(2)^2 + v_y(2)^2} = 20$ Substitute: $\sqrt{\left(\frac{v_0}{\sqrt{2}}\right)^2 + \left(\frac{v_0}{\sqrt{2}} - 20\right)^2} = 20$

Step 3: Solve for $v_0$

Square both sides: $\left(\frac{v_0}{\sqrt{2}}\right)^2 + \left(\frac{v_0}{\sqrt{2}} - 20\right)^2 = 400$ Simplify: $\frac{v_0^2}{2} + \left(\frac{v_0}{\sqrt{2}} - 20\right)^2 = 400$ Expand the second term: $\frac{v_0^2}{2} + \frac{v_0^2}{2} - 2 \cdot \frac{v_0}{\sqrt{2}} \cdot 20 + 400 = 400$ Simplify: $v_0^2 - \frac{40 v_0}{\sqrt{2}} + 400 = 400 \Rightarrow v_0^2 - \frac{40 v_0}{\sqrt{2}} = 0 \Rightarrow v_0 \left(v_0 - \frac{40}{\sqrt{2}}\right) = 0$ Since $v_0 \neq 0$, we get: $v_0 = \frac{40}{\sqrt{2}} = 20\sqrt{2} \text{ ms}^{-1}$

Step 4: Compute maximum height

The maximum height is reached when the vertical velocity becomes zero. Using: $H = \frac{v_{0y}^2}{2g} = \frac{\left(\frac{v_0}{\sqrt{2}}\right)^2}{2 \cdot 10} = \frac{v_0^2}{40}$ Substitute $v_0 = 20\sqrt{2}$: $H = \frac{(20\sqrt{2})^2}{40} = \frac{800}{40} = 20 \text{ m}$

Common Traps & Exam Tip:

• Ignoring vector nature of velocity: Students often forget that the velocity at any time is the vector sum of horizontal and vertical components. They may incorrectly assume the given speed is only vertical or horizontal.
• Sign errors in vertical velocity: The vertical velocity decreases with time due to gravity. A common mistake is using $+gt$ instead of $-gt$.
• Misapplying maximum height formula: Some students use $H = \frac{v_0^2}{2g}$ instead of $H = \frac{v_{0y}^2}{2g}$, forgetting to use the vertical component of initial velocity.
• Algebraic errors in solving for $v_0$: Expanding and simplifying the squared terms can be error-prone. Double-check each step to avoid sign or arithmetic mistakes.

Exam Tip: Always resolve the motion into horizontal and vertical components first. Use vector addition for velocity and remember that horizontal motion is uniform while vertical motion is uniformly accelerated.