JEE PYQ: Motion in a Plane - Question ID a9dc560eb0ac (JEE Main 2021)

In projectile motion, an object 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 (ignoring air resistance).
• Vertical motion: Motion under constant acceleration due to gravity ($g$), directed downward.

Key formulas used:

• Initial velocity components: $v_{0x} = v_0 \cos \theta, \quad v_{0y} = v_0 \sin \theta$
• Time to reach maximum height ($T$): At the highest point, the vertical component of velocity becomes zero. $v_y = v_{0y} - gT = 0 \implies T = \frac{v_{0y}}{g}$
• Maximum height ($h_{\text{max}}$): Using the equation of motion for vertical displacement. $h_{\text{max}} = v_{0y} T - \frac{1}{2} g T^2$ Alternatively, using the energy-like relation: $h_{\text{max}} = \frac{v_{0y}^2}{2g}$

Given:

• Initial speed, $v_0 = 25\ \text{ms}^{-1}$
• Launch angle, $\theta = 45^\circ$
• Acceleration due to gravity, $g = 10\ \text{ms}^{-2}$

Step 1: Resolve the initial velocity into components

The vertical component of the initial velocity is: $v_{0y} = v_0 \sin \theta = 25 \cdot \sin 45^\circ = 25 \cdot \frac{\sqrt{2}}{2} = 25 \cdot 0.7071 = 17.6775\ \text{ms}^{-1}$

Step 2: Calculate the time to reach maximum height ($T$)

At the highest point, the vertical velocity becomes zero: $v_y = v_{0y} - gT = 0 \implies T = \frac{v_{0y}}{g} = \frac{17.6775}{10} = 1.76775\ \text{s}$ Rounding to two decimal places, $T \approx 1.77\ \text{s}$

Step 3: Calculate the maximum height ($h_{\text{max}}$)

Using the formula: $h_{\text{max}} = \frac{v_{0y}^2}{2g} = \frac{(17.6775)^2}{2 \cdot 10} = \frac{312.5}{20} = 15.625\ \text{m}$ Alternatively, using the displacement equation: $h_{\text{max}} = v_{0y} T - \frac{1}{2} g T^2 = 17.6775 \cdot 1.76775 - \frac{1}{2} \cdot 10 \cdot (1.76775)^2$ $= 31.25 - 15.625 = 15.625\ \text{m}$

Step 4: Match with the given options

We have: $h_{\text{max}} = 15.625\ \text{m}, \quad T = 1.77\ \text{s}$ This matches Option C.

Common mistakes students make:

• Incorrect angle resolution: Some students confuse $\sin$ and $\cos$, leading to wrong velocity components. Always remember: $\sin$ for vertical, $\cos$ for horizontal.
• Using total velocity instead of vertical component: Students sometimes use $v_0$ directly in the height formula instead of $v_{0y}$, leading to incorrect $h_{\text{max}}$.
• Sign errors in acceleration: Gravity acts downward, so acceleration is $-g$ in upward motion. Forgetting the sign can lead to wrong time calculations.
• Rounding errors: Intermediate rounding (e.g., using $v_{0y} = 17.7$ instead of $17.6775$) can slightly alter the final answer. Use full precision until the final step.

Exam Tip: Always sketch the trajectory and label the components. This helps visualize the problem and avoid confusion between horizontal and vertical motions.